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[edit] Contents

Algebra

  • Geometry
  • Trigonometry
  • Introduction (page 1)
  • In simple terms
  • Radian and Degree Measure
  • The Unit Circle
  • Trigonometric Angular Functions
  • Right Angle Trigonometry
  • Graphs of Sine and Cosine Functions
  • Graphs of Other Trigonometric Functions
  • Inverse Trigonometric Functions
  • Applications and Models
  • Analytic Trigonometry
  • Using Fundamental Identities
  • Verifying Trigonometric Identities
  • Solving Trigonometric Equations
  • Sum and Difference Formulas
  • Multiple-Angle and Product-to-sum Formulas
  • Additional Topics in Trigonometry
  • Law of Sines
  • Law of Cosines
  • Vectors in the Plane
  • Vectors and Dot Products
  • Trigonometric Form of the Complex Number
  • Trigonometric Unit Circle and Graph Reference
  • Trigonometric Formula Reference
  • Trigonometric Identities Reference
  • Natural Trigonometric Functions of Primary Angles

[edit] Prerequisites And Basics

To be able to study Trigonometry successfully, it is recommended that students complete Geometry and Algebra prior to digging in to the course material. Students should also be familiar with the arithmetic of the real number system. It is helpful to have a graphing calculator and graph paper on hand to be able to follow along as well. If one is not available, software available on sites such as GraphCalc or GeoGebra may be helpful. Geometric constructions proposed in the text can be drawn using Geops, free software for performing geometric constructions in the manner of the Ancient Greeks.

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[edit] Trigonometry

[edit] Introduction

[edit] Introduction to trigonometry

A painting of the famous greek geometrist, and "father of measurement", Euclid. In the times of the greeks, trigonometry and geometry were important mathematical principles used in building, agriculture and education.

It is easy to explain in word-terms what trigonometry means, but it is more important to understand what mechanisms of thinking, which will help you understand not only trigonometry, but everything in life in a much more vivid way.

In mathematics, trigonometry is an important set of disciplines which relate to two and three dimensional objects; practically anything that you can see around you can be related to the principles of trigonometry and algebra -- in the real-world, it is very useful in engineering and construction, where its principles are important in accurately determining the lengths, sizes and areas of objects without having to actually create them first. Imagine the need to build a structure with only the basic land-area given to you: using the principles of trigonometry, you can easily calculate the geometric properties of objects to an unerring degree of accuracy.

Trigonometry, however, isn't just about using formulae to find the correct angle or size in school. It describes the relationships that occur naturally between objects and their similarity in structure. When we compare them using a similar set of ideas, it gives us a lot of power to understand the basis of other things in life beyond that of just their appearance. Even though we can look at a circle, an oval, square or rectangle, we can know that there are principles we can apply to their shape which can be expressed through one entity: the triangle.

[edit] A brief history of trigonometry

The moon; the natural circle which was thought to be the spark which started the determination of the nature of the circle, the angle, and trigonometry.

Originally, the Babylonians were the first to discover the measures of the angle, but it was not until the onset of the Greeks, who were the original pioneers in the field of trigonometry, and the inventors of a measure known as the "sexagesimal". In 2BC, a Greek man known as Hipparchus was thought to be the first person who devised a more complete idea of a trigonometric triangle. He produced a table of reference for solving a triangle's lengths and angles, by making a reference table of the lengths of the sides of the triangles for angles between 71o and 180o. This was what could be called the equivalent of a "sine table"; the basis of the modern sin function, which has become a crucial tool in the calculations for modern living, construction and manufacture. Sine tables were once used in some school systems in Europe and America, but have now been dropped for the use of the sin function on modern calculators, opting to focus more on the principles of trigonometry, rather than trigonometric values.

In his research, however, a crucial entity in recording was either lost, or not recorded simply because it may not have been thought of as important, or even thought of at all. This entity was known as the radius; half the length of the width of the circle, when measured from one side to another. Over 300 years later, Greeks adapted upon this measure by using the sexagesimal measure, and saying that the radius should be a fixed length of 60; r = 60.

An important figure, too, in geometry and trigonometry was Euclid. Euclid was a Greek mathematician, about whom almost nothing was known other than the works that he produced. Surprisingly, the work he is most renowned for, Elements, is an amazingly in-depth work for the time, as it covers in some detail the basic and more advanced aspects of geometry and trigonometry. Even though there is some uncertainty towards the originality of certain concepts contained within Elements, there is no doubt that Euclid is a very important figure in the discovery of trigonometric principles, because so much of what is known about geometric measure in trigonometry is comparable to his work.

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[edit] In simple terms

This page is intended as a simplified introduction to trigonometry. (This article is not always correctly formulated in mathematical language.)

[edit] Simple introduction

[edit] Introduction to Angles

If you are unfamiliar with angles, where they come from, and why they are actually required, this section will help you develop your understanding.

An angle between two lines in a 2-dimensional plane

In two dimensional flat space, two straight lines that are not parallel lines meet at an angle. Suppose you wish to measure the angle between two lines exactly so that you can tell a remote friend about it: draw a circle with its center located at the meeting of the two lines, making sure that the circle is small enough to cross both lines, but large enough for you to measure the distance along the circle's edge, the circumference, between the two cross points. Obviously this distance depends on the size of the circle, but as long as you tell your friend both the radius of the circle used, and the length along the circumference, then your friend will be able to reconstruct the angle exactly.

[edit] Triangle Definition

We know that triangles have three sides: each side is a line: if we choose any two sides of the triangle, then we have chosen two lines, which are not parallel, and must therefore meet at an angle. There are three ways that we can pick the two sides, so a triangle must have three angles: hence tri-angles, shortened to triangle.

This argument does not always work in reverse though. If you give me three angles, for example three right angles, I cannot make a triangle from them. This is also a problem with sides: I can give you three lengths that do not make the sides of a triangle: your height, the height of the nearest tree, the distance from the top of the tree to the center of the sun.

[edit] Triangle Ratios

Angles are not affected by the length of lines: an angle is invariant under transformations of scale (if you make the triangle bigger, its angles won't change). Given the three angles of a known triangle, you can draw a new triangle that is similar to the old triangle but not necessarily congruent. Two triangles are called similar if they have the same angles as each other, they are congruent if they have the same size angles and matching sides have the same length. Given a pair of similar triangles, if the lengths of two matching sides are equal, then the lengths of the other sides are also equal, and so the triangles are congruent. More generally, the ratios of the lengths of matching pairs of sides between two similar triangles are equal. In particular, a triangle can be divided into 4 congruent triangles by connecting the mid-points on the sides of the original triangle together, each congruent triangle is similar on half scale to the original triangle.

Consequently, angles are useful for making comparisons between similar triangles.

[edit] Right Angles

An angle of particular significance is the right-angle: the angle at each corner of a square or a rectangle. Indeed, a rectangle can always be divided into two triangles by drawing a line through opposing corners of the rectangle. This pair of triangles has interesting properties. First, the triangles are congruent. Second, they each have one angle which is a right-angle.

Imagine sitting at a table drawing a rectangle on a sheet of paper, and then dividing the rectangle into a pair of triangles by drawing a line from one corner to another. Perhaps the two triangles are different? A friend could sit on the other side of the table and draw the exact same thing by re-using your lines. The triangle nearest you has been produced by the exact same process as the triangle nearest your friend, therefore the two triangles are congruent, that is, identical. Try cutting a rectangle into two triangles to check.

Each triangle of this pair of triangles has one right angle: the rectangle has four right-angles, we split two of them by drawing from corner to corner, the remaining two were distributed equally to the identical triangles, so each triangle got one right-angle.

Thus every right angled triangle is half a rectangle.

A rectangle has four sides of two different lengths: two long sides and two short sides. When we split the rectangle into two identical right angled triangles, each triangle has a long side and a short side from the rectangle as well as a copy of the split line, which has a Greek name: "hypotenuse". In an aside, trigonometry has been defined as 'Many cheerful facts about the square of the hypotenuse'.

So the area of a right angled triangle is half the area of the rectangle from which it was split. Looking at a right angled triangle we can tell what the long and short sides of that rectangle were, they are the sides, the lines, that meet at a right angle. The area of the rectangle is the long side times the short side. The area of a right angled triangle is therefore half as much.

[edit] Right Triangles and Measurement

How to bisect an angle

It is possible to bisect any angle using only circles (which can be drawn with a compass) and straight lines by the following procedure:

  1. Call the vertex of the angle O. Draw a circle centered at O.
  2. Mark where the circle intersects each ray. Call these points A and B.
  3. Draw circles centered at A and B with equal radii, but make sure that these radii are large enough to make the circles intersect at two points. One sure way to do this is to draw line segment AB and make the radius of the circles equal to the length of that line segment. On the diagram, circles A and B are shown as near-half portions of a circle.
  4. Mark where these circles intersect, and connect these two points with a line. This line bisects the original angle.

A proof that the line bisects the angle is found in Proposition 9 of Book 1 of the Elements.

Given a right angle, we can use this process to split that right angle indefinitely to form any binary fraction (i.e., \frac{1}{2^n}, n \in \mathbb{Z}, e.g. \frac{1}{2}, \frac{1}{4}, \frac{1}{8}, \frac{1}{16}, \frac{1}{32}...) of it. Thus, we can measure any angle in terms of right angles. That is, a measurement system in which the size of the right angle is considered to be one, just as some believe that, with the distance measurement of feet, the length of King Henry I's foot was considered to be one.

[edit] Introduction to Radian Measure

En Radian cropped color.png
Of course, it is equally possible to start with a different sized angle and split it into two equal angles. We might consider the angle made by half a circle to be "one", or the angle made by a full circle to be "one".

Trigonometry is simplified if we choose the following strange angle as "one":

    1. Draw a circle with the radius marked.
    2. From the point where the radius meets the circumference of the circle, measure one radius length along the circumference of the circle moving counterclockwise and mark this point.
    3. Draw a line from this mark back to the center of the circle.
    4. The angle so formed is considered to be of size one in trigonometry; in order to tell your friends that this is the method you were using, you would say 'measuring in radians'.

[edit] Understanding the three sides of a Radian

To illustrate how the three sides of a radian relate to one another try the below thought experiment:

    1. Assume you have a piece of string that is exactly the length of the radius of a circle.
    2. Assume you have drawn a radian in the same circle. The radian has 3 points. One is the center of your circle and the other two are on the circumference of the circle where the sides of the radian intersect with the circle.
    3. Attach one end of the string at one of the points where the radian intersects with the circumference of the circle.
    4. Take the other end of the string and, starting at the point you chose in Step 2 above, trace the circumference of the circle towards the second point where the radian intersects with the circle until the string is pulled tight.
    5. You will see that the end of the string travels past the second point.
    6. This is because the string is now in a straight line. However, the radian has an arc for its third side, not a straight line. Even though a radian has three equal sides, the arc's curve causes the two points where the radian intersects with the circle to be closer together than they are to the third point of the triangle, which is at the center of our circle in this example.
    7. Now, with the string still pulled tight, find the half way point of the string, then pull it onto the circumference of the circle while allowing the end of the string to move along the circle's circumference. The end of the string is now closer to the second point because the path of the string is closer to the path of the circle's circumference.
    8. We can keep improving the fit of the string's path to the path of the circle's circumference by dividing each new section of the string in half and pulling it towards the circumference of the circle.

[edit] Is a radian affected by the size of its circle?

Does it matter what size circle is used to measure in radians? Perhaps using the radius of a large circle will produce a different angle than that produced by the radius of a small circle. The answer is no.

Recall our radian and circle from the experiment in the subsection above. Draw another circle inside the first circle, with the same center, but with half the radius. You will see that you have created a new radian inside the smaller circle that shares the same angle as the radian in the larger circle. We know that the two sides of the radians emanating from the center of the circles are equal to the radius of their respective circle. We also know that the third side of the radian in the larger circle (the arc) is also equal to the larger circle's radius. But how do we know that the third side of the radian in the smaller circle (the arc that follows the circumference of the smaller circle) is equal to the radius of the smaller circle? To see why we do know that the third side of the smaller circle's radian is equal to its radius, we first connect the two points of each radian that intersect with the circle with each other. By doing so, you will have created two isosceles triangles (triangles with two equal sides and two equal angles). I think you see where this is going.


An isosceles triangle has two equal angles and two equal sides. If you know one angle of any isosceles triangle and the length of two sides that make up that angle, then you can easily deduce the remaining characteristics of the isosceles triangle. For instance, if the two equidistant sides of an isosceles triangle intersect to form an angle that is equal to 40o, then you know the remaining angles must both equal 70 o. Since we know that the equidistant sides of our two isosceles triangles make up our known angle, then we can deduce that both of our radians (when converted to isosceles triangles with straight lines) have identical second and third angles. We also know that triangles with identical angles, regardless of their size, will have the length of their sides in a constant ratio to each other. For instance, we can deduce that an isosceles triangle will have sides that measure 2 meters by 4 meters by 4 meters if we know that an isoscleses triangle with identical angles measures 1 meter by 2 meters by 2 meters. Therefore, in our example, our isosceles triangle formed by the second smaller circle will have a third side exaclty equal to half of the third side of the isosceles triangle created by the larger circle. The relationship between the size of the sides of two triangles that share identical angles is also found in the relationship between radius and circumference of two circles that share the same center point - they will share the exact same ratio. In our example then, since the radius of our second smaller circle is exactly half of the larger circle, the circumference between the two points where the radian of the smaller circle intersect (which we have shown is one half of the distance between the two similar points on the larger circle) will share the same exact ratio. And there you have it - the size of the circle does not matter.

[edit] Using Radians to Measure Angles

Once we have an angle of one radian, we can chop it up into binary fractions as we did with the right angle to get a vast range of known angles with which to measure unknown angles. A protractor is a device which uses this technique to measure angles approximately. To measure an angle with a protractor: place the marked center of the protractor on the corner of the angle to be measured, align the right hand zero radian line with one line of the angle, and read off where the other line of the angle crosses the edge of the protractor. A protractor is often transparent with angle lines drawn on it to help you measure angles made with short lines: this is allowed because angles do not depend on the length of the lines from which they are made.

If we agree to measure angles in radians, it would be useful to know the size of some easily defined angles. We could of course simply draw the angles and then measure them very accurately, though still approximately, with a protractor: however, we would then be doing physics, not mathematics.

The ratio of the length of the circumference of a circle to its radius is defined as 2π, where π is an invariant independent of the size of the circle by the argument above. Hence if we were to move 2π radii around the circumference of a circle from a given point on the circumference of that circle, we would arrive back at the starting point. We have to conclude that the size of the angle made by one circuit around the circumference of a circle is 2π radians. Likewise a half circuit around a circle would be π radians. Imagine folding a circle in half along an axis of symmetry: the resulting crease will be a diameter, a straight line through the center of the circle. Hence a straight line has an angle of size π radians.

The angles of a triangle add up to a half circle angle, that is, an angle of size π radians. To see this, draw any triangle, mark the midpoints of each side and connect them with lines to subdivide the original triangles into 4 similar copies at half scale. A copy of each of the original triangles' three interior angles are juxtapositioned at each mid point, demonstrating that for any triangle the interior angles sum to a straight line, that is an angle of size π radians. This is true in particular for right angled triangles, which always have one angle of size π/2 radians, therefore the other two angles must also sum to a total size of π - π/2 = π/2 radians. If two sides of a right angled triangle have equal lengths, i.e. it is an isoceles right angled triangle, then each of the two other angles must be of size 1/2*π/2 = π/4 radians.

Folding a half circle in half again produces a quarter circle which must therefore have an angle of size π/2 radians. Is a quarter circle a right angle? To see that it is: draw a square whose corner points lie on the circumference of a circle. Draw the diagonal lines that connect opposing corners of the square, by symmetry they will pass through the center of the circle, to produce 4 similar triangles. Each such triangle is isoceles, and has an angle of size 2π/4 = π/2 radians where the two equal length sides meet at the center of the circle. Thus the other two angles of the triangles must be equal and sum to π/2 radians, that is each angle must be of size π/4 radians. We know that such a triangle is right angled, we must conclude that an angle of size π/2 radians is indeed a right angle.

[edit] Interior Angles of Regular Polygons

To demonstrate that a square can be drawn so that each of its four corners lies on the circumference of a single circle: Draw a square and then draw its diagonals, calling the point at which they cross the center of the square. The center of the square is (by symmetry) the same distance from each corner. Consequently a circle whose center is co-incident with the center of the square can be drawn through the corners of the square.

A similar argument can be used to find the interior angles of any regular polygon. Consider a polygon of n sides. It will have n corners, through which a circle can be drawn. Draw a line from each corner to the center of the circle so that n equal apex angled triangles meet at the center, each such triangle must have an apex angle of 2π/n radians. Each such triangle is isoceles, so its other angles are equal and sum to π - 2π/n radians, that is each other angle is (π - 2π/n)/2 radians. Each corner angle of the polygon is split in two to form one of these other angles, so each corner of the polygon has 2*(π - 2π/n)/2 radians, that is π - 2π/n radians.

This formula predicts that a square, where the number of sides n is 4, will have interior angles of π - 2π/4 = π - π/2 = π/2 radians, which agrees with the calculation above.

Likewise, an equilateral triangle with 3 equal sides will have interior angles of π - 2π/3 = π/3 radians.

A hexagon will have interior angles of π - 2π/6 = 4π/6 = 2π/3 radians which is twice that of an equilateral triangle: thus the hexagon is divided into equilateral triangles by the splitting process described above.

[edit] Summary and Extra Notes

In summary: it is possible to make deductions about the sizes of angles in certain special conditions using geometrical arguments. However, in general, geometry alone is not powerful enough to determine the size of unknown angles for any arbitrary triangle. To solve such problems we will need the help of trigonometric functions.

In principle, all angles and trigonometric functions are defined on the unit circle. The term unit in mathematics applies to a single measure of any length. We will later apply the principles gleaned from unit measures to larger (or smaller) scaled problems. All the functions we need can be derived from a triangle inscribed in the unit circle: it happens to be a right-angled triangle.
The hypotenuse rotate counter-clockwise

The center point of the unit circle will be set on a Cartesian plane, with the circle's centre at the origin of the plane — the point (0,0). Thus our circle will be divided into four sections, or quadrants.

Quadrants are always counted counter-clockwise, as is the default rotation of angular velocity ω (omega). Now we inscribe a triangle in the first quadrant (that is, where the x- and y-axes are assigned positive values) and let one leg of the angle be on the x-axis and the other be parallel to the y-axis. (Just look at the illustration for clarification). Now we let the hypotenuse (which is always 1, the radius of our unit circle) rotate counter-clockwise. You will notice that a new triangle is formed as we move into a new quadrant, not only because the sum of a triangle's angles cannot be beyond 180°, but also because there is no way on a two-dimensional plane to imagine otherwise.

[edit] Angle-values simplified

Imagine the angle to be nothing more than exactly the size of the triangle leg that resides on the x-axis (the cosine). So for any given triangle inscribed in the unit circle we would have an angle whose value is the distance of the triangle-leg on the x-axis. Although this would be possible in principle, it is much nicer to have a independent variable, let's call it phi, which does not change sign during the change from one quadrant into another and is easier to handle (that means it is not necessarily always a decimal number).

!!Notice that all sizes and therefore angles in the triangle are mutually directly proportional. So for instance if the x-leg of the triangle is short the y-leg gets long.

That is all nice and well, but how do we get the actual length then of the various legs of the triangle? By using translation tables, represented by a function (therefore arbitrary interpolation is possible) that can be composed by algorithms such as Taylor. Those translation-table-functions (sometimes referred to as LUT, Look up tables) are well known to everyone and are known as sine, cosine and so on. (Whereas of course all the abovementioned latter ones can easily be calculated by using the sine and cosine).

In fact in history when there weren't such nifty calculators available, printed sine and cosine tables had to be used, and for those who needed interpolated data of arbitrary accuracy - taylor was the choice of word.

So how can I apply my knowledge now to a circle of any scale. Just multiply the scaling coefficient with the result of the trigonometric function (which is referring to the unit circle).

And this is also why cos(φ)2 + sin(φ)2 = 1, which is really nothing more than a veiled version of the pythagorean theorem: cos(φ) = a;sin(φ) = b;a2 + b2 = c2, whereas the c = 12 = 1, a peculiarity of most unit constructs. Now you also see why it is so comfortable to use all those mathematical unit-circles.

Another way to interprete an angle-value would be: A angle is nothing more than a translated 'directed'-length into which the information of the actual quadrant is packed and the applied type of trigonometric function along with its sign determines the axis ('direction'). Thus something like the translation of a (x,y)-tuple into polar coordinates is a piece of cake. However due to the fact that information such as the actual quadrant is 'translated' from the sign of x and y into the angular value (a multitude of 90) calculations such as for instance the division in polar-form isn't equal to the steps taken in the non-polar form.

Oh and watch out to set the right signs in regard to the number of quadrant in which your triangle is located. (But you'll figure that out easily by yourself).

I hope the magic behind angles and trigonometric functions has disappeared entirely by now, and will let you enjoy a more in-depth study with the text underneath as your personal tutor.


Next Page: Radians and Degrees
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[edit] Radian and Degree Measure

[edit] A Definition and Terminology of Angles

An angle is determined by rotating a ray about its endpoint. The starting position of the ray is called the initial side of the angle. The ending position of the ray is called the terminal side. The endpoint of the ray is called its vertex. Positive angles are generated by counter-clockwise rotation. Negative angles are generated by clockwise rotation. Consequently an angle has four parts: its vertex, its initial side, its terminal side, and its rotation.

An angle is said to be in standard position when it is drawn in a cartesian coordinate system in such a way that its vertex is at the origin and its initial side is the positive x-axis.

an angle in standard position

[edit] The radian measure

One way to measure angles is in radians. To signify that a given angle is in radians, a superscript c, or the abbreviation rad might be used. If no unit is given on an angle measure, the angle is assumed to be in radians.

\frac{3\pi}{2}^{c} \equiv \frac{3\pi}{2} \;\mathrm{rad.} \equiv \frac{3\pi}{2}

[edit] Defining a radian

A single radian is defined as the angle formed in the minor sector of a circle, where the minor arc length is the same as the radius of the circle.

Defining a radian with respect to the unit circle.


1 \approx 57.296^{\circ}

[edit] Measuring an angle in radians

The size of an angle, in radians, is the length of the circle arc s divided by the circle radius r.

\mbox{angle in radians} = \frac sr

Measuring an Angle in Radians

We know the circumference of a circle to be equal to r, and it follows that a central angle of one full counterclockwise revolution gives an arc length (or circumference) of s = 2πr. Thus 2 π radians corresponds to 360°, that is, there are radians in a circle.

[edit] Converting from Radians to Degrees

Because there are 2π radians in a circle:

To convert degrees to radians:
 \theta ^{c} = \theta^\circ \times \frac{\pi}{180}

To convert radians to degrees:
 \phi ^\circ= \phi^{c} \times \frac{180}{\pi}

[edit] Exercises

[edit] Exercise 1:Degree-Radian Conversion

[edit] Exercise 2: Radian-Degree Conversion

Convert the following angle measurements from radians to degrees.



Next Page: The Unit Circle
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[edit] The Unit Circle

The Unit Circle is a circle with its center at the origin (0,0) and a radius of one unit.

Unit Circle

Angles are always measured from the positive x-axis (also called the "right horizon"). Angles measured counterclockwise have positive values; angles measured clockwise have negative values.

A unit circle with certain exact values marked on it is below:

Labeled Unit Circle
.

It is well worth the effort to memorize the values of sine and cosine on the unit circle (cosine is equal to x while sine is equal to y) included in this unit circle, since the information will be useful for many math concepts.

Unit circles form the basis of most analog clocks and animations on computers since the cos and sin correspond to the x and y positions of the end of the line segments representing the hands of the clock. The angle of a hand is calculated (-6° or -π/30 radians per hour, minute or second past the 3 position on the clock, depending on the hand) and the cosine and sine of the angle are used to calculate the co-ordinates of the end point of the hand.

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[edit] Trigonometric Angular Functions

[edit] Geometrically defining sine and cosine

In the unit circle shown here, a unit-length radius has been drawn from the origin to a point (x,y) on the circle.

Defining sine and cosine

A line perpendicular to the x-axis, drawn through the point (x,y), intersects the x-axis at the point with the abscissa x. Similarly, a line perpendicular to the y-axis intersects the y-axis at the point with the ordinate y. The angle between the x-axis and the radius is α.

We define the basic trigonometric functions of any angle α as follows:


\begin{matrix}
\mathrm{Sine:} & \sin(\alpha) & = & y \\
\mathrm{Cosine:} & \cos(\alpha) & = & x \\
\end{matrix}

tanθ can be algebraically defined.

\tan\theta = \frac{\sin\theta}{\cos\theta} \qquad \cos\theta \ne 0

\tan\alpha = \frac{y}{x} \qquad x \ne 0

These three trigonometric functions can be used whether the angle is measured in degrees or radians as long as it specified which, when calculating trigonometric functions from angles or vice versa.

[edit] Geometrically defining tangent

In the previous section, we algebraically defined tangent, and this is the definition that we will use most in the future. It can, however, be helpful to understand the tangent function from a geometric perspective.

Geometrically defining tangent

A line is drawn at a tangent to the circle: x = 1. Another line is drawn from the point on the radius of the circle where the given angle falls, through the origin(O), to a point on the drawn tangent(Q). The ordinate of this point is called the tangent of the angle.

The slope of the line OQ = \frac {KC}{OC} and as we mentioned before

KC = sin(θ) , OC = cos(θ)

Hence , the Slope of the line OQ = \frac {\sin(\theta)}{\cos(\theta)}

and also the slope of OQ = \frac {QP}{OP} = \frac {QP}{1} = \frac {\tan(\theta)}{1} = tan(θ)

Hence , we can deduce that tan(θ) = \frac {KC}{OC} = \frac {\sin(\theta)}{\cos(\theta)} = QP = the ordinate of the point Q = the slope of OQ

[edit] Domain and range of circular functions

Any size angle can be the input to sine or cosine — the result will be as if the largest multiple of 2π (or 360°) were subtracted from the angle. The output of the two functions is limited by the absolute value of the radius of the unit circle, | 1 | .


\begin{matrix}
 & \mathrm{domain}&\mathrm{range} \\
\mathrm{sine} & \mathbb{R} & [-1,1] \\
\mathrm{cosine} & \mathbb{R} & [-1,1] \\
\end{matrix}

R represents the set of all real numbers.

No such restrictions apply to the tangent, however, as can be seen in the diagram in the preceding section. The only restriction on the domain of tangent is that odd integer multiples of \frac{\pi}{2} are undefined, as a line parallel to the tangent will never intersect it.


\begin{matrix}
 & \mathrm{domain}&\mathrm{range} \\
\mathrm{tangent} & \mathbb{R} \setminus \left \{ \cdots,-\frac{\pi}{2},\frac{\pi}{2},\frac{3\pi}{2},\frac{5\pi}{2},\cdots \right \} & \mathbb{R} \\
\end{matrix}

for a deep understanding of trigonometric functions explore this Applet

[edit] Applying the trigonometric functions to a right-angled triangle

If you redefine the variables as follows to correspond to the sides of a right triangle:
• x = a (adjacent)
• y = o (opposite)
• a = h (hypotenuse)


Next Page: Right Angle Trigonometry
Previous Page: The unit circle

Home: Trigonometry

[edit] Right Angle Trigonometry


[edit] Construction of Right Triangles

Right triangles are easily constructed. Recall that a diameter is a straight line which starts at one point on the circle and goes through the center to the other side. Using this property of a circle:

  1. Construct a diameter of a circle. Call the points where the diameter touches the circle a and b.
  2. Choose a distinct point c on the circle (any point that is neither a nor b).
  3. Construct line segments connecting points a, b and c.

Provided the above three directions are followed, the resulting triangle Δabc will be a right triangle. Note that the right angle will be formed on the circle unless intentionally designated as the single instance at which it exists at the center point (e.g. when the circle is divided into equal quadrants by two perpendicular diameters).

This right triangle can be further divided into two isosceles triangles by adding a line segment from c to the center of the circle.

To simplify the following discussion, we specify that the circle has diameter 1 and is oriented such that the diameter drawn above runs left to right. We shall denote the angle of the triangle at the right θ and that at the left φ. The sides of the right triangle will be labeled, starting on the right, a, b, c, so that a is the rightmost side, b the leftmost, and c the diameter. We know from earlier that side c is opposite the right angle and is called the hypotenuse. Side a is opposite angle φ, while side b is opposite angle θ. We reiterate that the diameter, now called c, shall be assumed to have length 1 except where otherwise noted.

By constructing a right angle to the diameter at one of the points where it crosses the circle and then using the method outlined earlier for producing binary fractions of the right angle, we can construct one of the angles, say θ, as an angle of known measure between 0 and π / 2. The measure of the other angle φ is then π - π/2 - θ = π/2 - θ, the complement of angle θ. Likewise, we can bisect the diameter of the circle to produce lengths which are binary fractions of the length of the diameter. Using a compass, a binary fractional length of the diameter can be used to construct side a (or b) having a known size (with regard to the diameter c) from which side b can be constructed.

[edit] Using Right Triangles to find Unknown Sides

[edit] Finding unknown sides from two other sides

From the Theorem of Pythagoras, we know that:

a^2 + b^2 = c^2. \,

Recall that c (the diameter/hypotenuse in the above example) has been defined as having a length of one, therefore:

b^2 = 1 - a^2. \,

permits us to calculate the length of b squared. It may happen that b squared is a fraction such as 1/4 for which a rational square root can be found, in this case as 1/2, alternatively, we could use Newton's Method to find an approximate value for b.

[edit] Finding unknown sides from a side and a (non-right) angle

As any triangle could be compared to our basic triangle (formed from a circle with a diameter of one), a table enumerating the relationships between angles and side lengths would be very useful to understanding the properties of any triangle. However, such a table would be unwieldy in practice and it is often not necessary to know the exact value.

Of course, given an angle θ, we could construct a right angled triangle using ruler and compass that had \theta\, as one of its angles, we could then measure the length of the side that corresponds to a to evaluate the cos function. Such measurements would necessarily be in-exact; it would be a problem in physics to see how accurately such measurements can be made; using trigonometry we can make precise predictions with which the results of these physical measurements can be compared.

The most common way of communicating the idea that relationships exist without providing exact details takes the form of a 'function'. A function is like a machine that takes some simple input and produces some simple output. Usually a function defines some kind of rule ([Function]) and provides us a handy notation useful in trigonometry. That way, we know that we're using a certain relationship without needing to know the exact numerical values. Basic trigonometric functions are simply stand-ins for the relationship between angles and sides of a triangle.

One such function, which allows us to know the relationship between any value of \theta\, and the corresponding value of a\, is called the cosine or \cos\,. This universal relationship is represented as: \frac{a}{c} = \cos\theta which would save us the work of constructing angles and lengths and making difficult deductions from them.

This means that if you know the cosine of an angle, you also know the relationship between the lengths of the sides. The actual size of the triangle can be bigger or smaller, but the mathematical relationship represented by the cosine does not change so long as the size of the angle remains constant.

[edit] Cosine example

Some explicit values for the cos function are known. For \theta\, = 0, sides a\, and c\, are coincidental: a = c = \frac{a}{c} = 1, so \cos\left(0\right) = 1. For \theta = \frac{\pi}{2}, sides b\, and c\, are coincidental and of length 1, and side a\, is of zero length, consequently a\, = 0, c\, = 1. \frac{a}{c} = 0 and \cos\frac{\pi}{2} = 0.


The simplest right angle triangle we can draw is the isosceles right angled triangle, it has a pair of angles of size \frac{\pi}{4} radians, and if its hypotenuse is considered to be of length one, then the sides a\, and b are of length \sqrt{\frac{1}{2}} as can be verified by the theorem of Pythagoras. If the side a\, is chosen to be the same length as the radius of the circle containing the right angled triangle, then the right hand isosceles triangle obtained by splitting the right angled triangle from the circumference of the circle to its center is an equilateral triangle, so θ must be \frac{\pi}{3}, and φ must be \frac{\pi}{6} and b\,^2 must be 1 - \frac{1}{2}\cdot\frac{1}{2} = \frac{3}{4}.

[edit] Properties of the cosine and sine functions

[edit] Period

Ancient observation of the circle (and its definition of perfect roundness) lead to the finding that circles increase in size by the value of 2π. Therefore, a perfect circle is maintained, along with the relationships formed by triangles within, by adding 2π to any angle θ. This is called the period, --the size of the angle or the time period over which the relationships begin to repeat (correlating the two is complex, and allows us to talk about wave theory).

Using functions, we can represent this fact in terms of the cosine function by stating that : \cos\theta = \cos\left(\theta + 2\pi n \right),\,

Knowing the period of the sine and cosine functions (and by derivation, that of other functions) is useful because it means that we can substitute one angle for another when we know that the period is the same. This helps in calculations, such as when there is the need to add or subtract angles.

[edit] Half Angle and Double Angle Formulas

We can derive a formula for cos(θ/2) in terms of cos(θ) which allows us to find the value of the cos() function for many more angles. To derive the formula, draw an isoceles triangle, draw a circle through its corners, connect the center of the circle with radii to each corner of the isoceles triangle, extend the radius through the apex of the isoceles triangle into a diameter of the circle and connect the point where the diameter crosses the other side of the circle with lines to the other corners of the isoceles triangle.

Therefore:

\cos (\theta /2) = \sqrt{\frac {1 + \cos \theta}{2}}

which gives a method of calculating the cos() of half an angle in terms of the cos() of the original angle. For this reason * is called the "Cosine Half Angle Formula".

The half angle formula can be applied to split the newly discovered angle which in turn can be split again ad infinitum. Of course, each new split involves finding the square root of a term with a square root, so this cannot be recommended as an effective procedure for computing values of the cos() function.

Equation * can be inverted to find cos(θ) in terms of cos(θ)/2:

\begin{matrix}&\cos (\theta /2)  =  \sqrt{\frac{1 + \cos (\theta)}{2}}\\
  \Rightarrow  &  \cos^2(\theta/2)  =  \frac{1 + \cos(\theta)}{2} \\
  \Rightarrow  &  \cos^2(\theta/2) - \frac{1}{2}  =  \frac{\cos (\theta)}{2}  \\
  \Rightarrow  & 2\cos^2(\theta/2) - 1   =  \cos(\theta) \\
 \end{matrix}

substituting \delta\ = \frac{\theta}{2} gives:

2\cos^2(\delta) - 1     = \cos(2\delta) \,

that is a formula for the cos() of double an angle in terms of the cos() of the original angle, and is called the "cosine double angle sum formula".

[edit] Angle Addition and Subtraction formulas

To find a formula for \cos(\theta_1+\theta_2)\, in terms of \cos\left( \theta_1\right) and \cos\left( \theta_2\right): construct two different right angle triangles each drawn with side c\, having the same length of one, but with \theta_1 \ne \theta_2, and therefore angle \psi_1 \ne \psi_2. Scale up triangle two so that side a_2\, is the same length as side c_1\,. Place the triangles so that side c_1\, is coincidental with side a_2\,, and the angles \theta_1\, and \theta_2\, are juxtaposed to form angle \theta_3\,=\theta_1\,+\theta_2\, at the origin. The circumference of the circle within which triangle two is embedded (circle 2) crosses side a_1\, at point g\,, allowing a third right angle to be drawn from angle \varphi_2 to point g\, . Now reset the scale of the entire figure so that side c_2\, is considered to be of length 1. Side a_2\, coincidental with side c_1\, will then be of length \cos\left( \theta_1\right), and so side a_1\, will be of length \cos\left( \theta_1\right) \cdot \cos\left( \theta_2\right) in which length lies point g\,. Draw a line parallel to line a_1\, through the right angle of triangle two to produce a fourth right angle triangle, this one embedded in triangle two. Triangle 4 is a scaled copy of triangle 1, because:

 (1) it is right angled, and 
 (2) \theta_4 + (\pi - \frac{\pi}{2} - \theta_1 - \theta_2) = \varphi_2 = \pi - \frac{\pi}{2} - \theta_2 \Rightarrow \theta_4 = \theta_1. 

The length of side b4 is cos(φ2)cos(φ4) = cos(φ2)cos(φ1) as θ4 = θ1 Thus point g is located at length:

 cos(θ1+θ2) = cos(θ1)cos(θ2) - cos(φ1)cos(φ2), where θ1+φ1 = θ2+φ2 = π/2

giving us the "Cosine Angle Sum Formula".

[edit] Proof that angle sum formula and double angle formula are consistent

We can apply this formula immediately to sum two equal angles:

   cos(2θ) = cos(θ+θ) = cos(θ)cos(θ) - cos(φ)cos(φ) where θ+φ = π/2
           = cos(θ)**2 - cos(φ)**2           (I)

From the theorem of Pythagoras we know that:

   a**2 + b**2 = c**2

in this case:

   cos(θ)**2 + cos(φ)**2  = 1**2                  where θ+φ = π/2
 =>            cos(φ)**2  = 1 - cos(θ)**2

Substituting into (I) gives:

   cos(2θ)   = cos(θ)**2 -      cos(φ)**2         where θ+φ = π/2 
             = cos(θ)**2 - (1 - cos(θ)**2)
             = 2 * cos(θ)**2 - 1

which is identical to the "Cosine Double Angle Sum Formula":

   2 * cos(δ)**2 - 1 = cos(2δ)

[edit] Derivative of the Cosine Function

We are now in a position to evaluate the expression:

 cos(θ+δθ)/ δθ = (cos(θ)cos(δθ) - cos(δθ)cos(δφ)) /δθ   where θ+φ = δθ+δφ = π/2.

If we let δθ tend to zero we get an increasingly accurate expression for the rate at which cos(θ) varies with θ. If we let δθ tend to zero, cos(δθ) will tend to one, δφ will tend to π/2, cos(δφ) will tend to zero, and cos(δφ)/δθ will tend to one: allowing us to write in the limit:

 (cos(θ+δθ) - cos(θ))/δθ = (cos(θ)cos(δθ)-cos(φ)cos(δφ)-cos(θ)) /δθ   where θ+φ = δθ+δφ = π/2.
                         = (cos(θ) * 1   -cos(φ) * δθ  -cos(θ)) /δθ 
                         = -cos(φ) * δθ /δθ 
                         = -cos(φ)                                    where θ+φ = π/2.
                         = -cos(π/2 - θ)

The function cos(π/2 - θ) occurs with such prevalence that it has been given a special name:

 sin(θ) = cos(π/2 - θ) 
        = cos(φ) where θ+φ = π/2.

"sin" is pronounced "sine".

The above result can now be stated more succinctly:

 (cos(θ+δθ) - cos(θ))/δθ = -sin(θ)

often further abbreviated to:

 d cos(θ)/δθ = -sin(θ)

or in words: the rate of chage of cos(θ) with θ is -sin(θ).

[edit] Pythagorean identity

Armed with this definition of the sin() function, we can restate the Theorem of Pythagoras for a right angled triangle with side c of length one, from:

    cos(θ)**2 + cos(φ)**2 = 1**2  where θ+φ = π/2

to:

    cos(θ)**2 + sin(θ)**2 = 1.

We can also restate the "Cosine Angle Sum Formula" from:

 cos(θ1+θ2) = cos(θ1)cos(θ2) - cos(φ1)cos(φ2), where θ1+φ1 = θ2+φ2 = π/2

to:

 cos(θ1+θ2) = cos(θ1)cos(θ2) - sin(θ1)sin(θ2)

[edit] Sine Formulas

The price we have to pay for the notational convenience of this new function sin() is that we now have to answer questions like: Is there a "Sine Angle Sum Formula". Such questions can always be answered by taking the cos() form and selectively replacing cos(θ)**2 by 1 - sin(θ)**2 and then using algebra to simplify the resulting equation. Applying this technique to the "Cosine Angle Sum Formula" produces:

   cos(θ1+θ2)        =      cos(θ1)cos(θ2) - sin(θ1)sin(θ2)
=> cos(θ1+θ2)**2     =     (cos(θ1)cos(θ2) - sin(θ1)sin(θ2))**2
=> 1 - cos(θ1+θ2)**2 = 1 - (cos(θ1)cos(θ2) - sin(θ1)sin(θ2))**2
=> sin(θ1+θ2)**2     = 1 - (cos(θ1)**2*cos(θ2)**2 + sin(θ1)**2*sin(θ2)**2 - 2*cos(θ1)cos(θ2)sin(θ1)sin(θ2))   -- Pythagoras on left, multiply out right hand side
                     = 1 - (cos(θ1)**2*(1-sin(θ2)**2) + sin(θ1)**2*(1-cos(θ2)**2) - 2*cos(θ1)cos(θ2)sin(θ1)sin(θ2)) -- Carefully selected Pythagoras again on the left hand side
                     = 1 - (cos(θ1)**2-cos(θ1)**2*sin(θ2)**2 + sin(θ1)**2-sin(θ1)**2*cos(θ2)**2 - 2*cos(θ1)cos(θ2)sin(θ1)sin(θ2)) -- Multiplied out
                     = 1 - (1         -cos(θ1)**2*sin(θ2)**2             -sin(θ1)**2*cos(θ2)**2 - 2*cos(θ1)cos(θ2)sin(θ1)sin(θ2)) -- Carefully selected Pythagoras 
                     =                 cos(θ1)**2*sin(θ2)**2              sin(θ1)**2*cos(θ2)**2 + 2*cos(θ1)cos(θ2)sin(θ1)sin(θ2)) -- Algebraic simplification
                     =                (cos(θ1)sin(θ2)                    +sin(θ1)cos(θ2))**2

taking the square root of both sides produces the the "Sine Angle Sum Formula"

=> sin(θ1+θ2) = cos(θ1)sin(θ2) + sin(θ1)*cos(θ2)

We can use a similar technique to find the "Sine Half Angle Formula" from the "Cosine Half Angle Formula":

cos(θ/2) = squareRoot(1/2 + cos(θ)/2).

We know that 1 - cos(θ/2)**2 = sin(θ/2) **2, so squaring both sides of the "Cosine Half Angle Formula" and subtracting from one:

 => (1 - cos(θ/2)**2) =       1 - (1/2 + cos(θ)/2)      
 =>      sin(θ/2)**2  =            1/2 - cos(θ)/2      
 =>      sin(θ/2)     = squareRoot(1/2 - cos(θ)/2)      

So far so good, but we still have a cos(θ/2) to get rid of. Use Pythagoras again to get the "Sine Half Angle Formula":

         sin(θ/2)     = squareRoot(1/2 - squareRoot(1 - sin(θ)**2)/2)  

or perhaps a little more legibly as:

         sin(θ/2)     = squareRoot(1/2)*squareRoot(1 - squareRoot(1 - sin(θ)**2))

[edit] On the Use of Computers for Algebraic Manipulation

The algebra to perform such manipulations tends to be horrendous: the horrors can be reduced by using automated symbolic algebraic manipulation software to confirm that each line of the derivation has been carried out correctly. However, such software is unable, in of itself, to perform the set of deductions above without human intervention: you will note that at various points in the above sequence, the Theorem of Pythagoras was cunningly applied for reasons that only became obvious towards the end of the derivation.

Alternatively, one can draw a plausible geometric diagram, make careful deductions about unknown sides and angles from known sides and angles. Again, software can check that each inference is consistent with the known facts, however, such software cannot decide which diagram to draw in the first place, nor which are the significant inferences to be made. You will no doubt have experienced the alarming feeling when manipulating such equations or diagrams that the numbers of terms in the equations, or the numbers of lines, sides and circles on the diagram, are doubling after each possible inference, multiplying exponentially beyond your control, with no solution in sight. Undirected software rapidly looses itself in this exponential maze despite the protestations of the salaried scientists held captive by Microsoft at Cambridge University, who protest, I think too much, that Alan Turing and Kurt Godell were wrong, and that "Machines can do Mathematics" if only they are given enough time to program (that is direct) them, to do so. Fortunately, the most that machines can do, is: assist humans with the tedious but necessary task of verifying each logical deduction. Machines cannot see the beauty of geometry, nor find directions to solve equations.

As is possible to define many other ratios of sides in an right angled triangle, the ratio of side b to side a for example. There will surely be relationships between these newly defined auxiliary trigonometric functions - it would be very odd if there were not. However, all such subsidiary functions can be dealt with by the properties of the cos() function alone; any newly defined functions must earn their keep by reducing complexity, not increasing it: their existence often creates more definitions to learn without generating any significantly new knowledge. Mathematics should be derivable from as few ideas as possible, that is from quality without quantity. Those of you, other than the physics students liberated by Feynmann in Brazil, who are facing exams produced at the behest of tyrannical governments will be reduced to the tedium of learning the many intricate relationships between these auxiliary functions, so that a bureaucrat with a fetish for metrics can kowtow to public opinion: please keep in mind that quantity and quality are not necessarily the same in mathematics; that increasing the number of auxiliary functions and their interrelationships merely aggravates the number of possibilities in each step of the solution space, without offering you any useful guidance as to which would be the most to use.

[edit] Sine derivative

We discovered earlier that the rate of change of cos(θ) with θ is -cos(φ) where θ+φ = π/2, or in the new fangled notation introduced above: -sin(θ). If we increase θ by δθ, we reduce φ by δθ, that is the rate of change of φ with θ is -1,

 => dφ / dθ = -1, => -dθ / dφ = 1.

We have also proved above that

 d cos(θ)/dθ = -sin(θ)

And we know from the definition that:

 cos(θ) = sin(φ)

Putting these knows together we can find the rate of change of sin(θ) with θ:

 d sin(θ)/dθ = d cos(φ)/dθ = d cos(φ)/dθ * -dθ/dφ = -1* d cos(φ)/dφ = -1* -cos(θ) = cos(θ)

that is the rate of change of sin(θ) with is cos(θ). To summarize:

d  sin(θ) / dθ =  cos(θ)
d  cos(θ) / dθ = -sin(θ)
d -sin(θ) / dθ = -cos(θ)
d -cos(θ) / dθ =  sin(θ)

We are going round in circles! Notice that we have to apply the rate of change operation 4 times to get back to where we started.

[edit] Relationships between exponential function and trigonometric functions

A similar function is the exponential function e(θ) which is defined by the statement: the rate of change of e(θ)is e(θ) and the value of e(0) is 1. Here we only have to apply the rate of change operator once to get back where we started. Explicitly:

 (e(θ+δθ)-e(θ)) / δθ tends to e(θ) as δθ tends to zero.

which can be rewritten replacing θ by iθ where i is any constant number, to get:

 (e(iθ+δiθ)-e(iθ)) / δiθ tends to e(iθ) as δθ tends to zero.

because i is a constant, δiθ = iδθ, performing this substitution, we get:

    (e(iθ+iδθ)-e(iθ))  / iδθ tends to e(iθ) as δθ tends to zero.
=>  (e(i(θ+δθ))-e(iθ)) / iδθ tends to e(iθ) as δθ tends to zero.
=>  d e(iθ) / dθ  = i*e(iθ)

That is, the rate of change of e(iθ) with θ is i*e(iθ). We can continue this process to find the rate of change of i*e(iθ) with θ:

  d i*e(iθ) / dθ                                    is the limit of: 
  (i* e(i(θ+δθ))-i*e(iθ)) / δθ as δθ tends to zero, which is the same as:
   i*(e(i(θ+δθ))-  e(iθ)) / δθ as δθ tends to zero, which is:
   i* d e(iθ) / dθ = i*i*e(iθ).

Performing this 4 time successively yields and comparing with the same action on the cos() function:

   d       e(iθ) / dθ  =       i*e(iθ)               d   cos(θ) / dθ  = - sin(θ)
   d     i*e(iθ) / dθ  =     i*i*e(iθ)               d - sin(θ) / dθ  = - cos(θ)
   d   i*i*e(iθ) / dθ  =   i*i*i*e(iθ)               d - cos(θ) / dθ  =   sin(θ)
   d i*i*i*e(iθ) / dθ  = i*i*i*i*e(iθ)               d   sin(θ) / dθ  =   cos(θ)

If only there was a number i, such that i*i = -1, and hence i*i * i*i = -1 * -1 = 1, then we could relate the function cos() to the function e(). Fortunately, there is a number that will work: the square roots of -1. From here on i will denote a square root of -1. -i*-i = (-1)*i*(-1)*i = (-1)*(-1)*i*i = (1)*(-1)=-1 is also a solution. We can expect then that cos(θ) is some linear combination of e(iθ) and e(-iθ), perhaps:

 cos(θ) = A*e(iθ)+B*e(-iθ)

We know that cos(0) = 1, so:

 cos(0) = 1 = A*e(i*0)+B*e(-i*0) = A*e(0)+B*e(0) = A + B

Finding the rate of change with θ:

=> d cos(θ) / dθ = d (A*e(iθ)) / dθ +d  (B*e(-iθ))/dθ 
=> - sin(θ)      =  i*A*e(iθ)       + -i*B*e(-iθ) = - cos(π/2-θ)

setting θ = 0, remembering that cos(π/2) = 0

=> 0 = iA - iB = i(A-B) => A = B

So now we know that A+B = 1 and A = B, so A+A = 1, so A = 1/2 and B = 1/2.

To summarize what we know so far:

cos(θ) = e(iθ)/2 + e(-iθ)/2   where θ is in radians and i is a square root of -1.

Replacing θ by -θ gives conversely:

cos(-θ) = e(-iθ)/2 + e(iθ)/2, the same formula, so we must have that cos(θ) =  cos(-θ).

Given:

cos(θ) = e(iθ)/2 + e(-iθ)/2 

We can find the rate of change with θ of both sides to fined the sin() in terms of e()

    -sin(θ)  =  i+e( iθ)/2 - i*e(-iθ)/2
 =>  sin(θ)  =  i*e(-iθ)/2 - i*e( iθ)/2

Substituting -θ for θ gives:

     sin(-θ)  =  i*e(iθ)/2 - i*e(-iθ)/2 = -sin(θ)

thus sin() is an odd function, compare this to cos() which is an even function because cos(θ) = cos(-θ).

We can find the function e() in terms of cos() and sin():

 cos(θ) + i * sin(θ)  =  e(iθ)/2 + e(-iθ)/2 + i*i*e(-iθ)/2 - i*i*e(iθ)/2
                      =  e(iθ)/2 + e(-iθ)/2 -     e(-iθ)/2 +     e(iθ)/2
                      =  e(iθ)

This is called "Euler's Formula".

From the "Cosine Double Angle Formula", we know that:

cos(2θ) = 2 * cos(θ)**2 - 1

Let θ = π/2, so that cos(π/2) = 0, then:

   cos(2π/2) = 2 * cos(π/2)**2 - 1
=> cos (π)   = 2 * 0**2        - 1
=> cos (π)   = - 1

By Pythagoras:

   sin(π)**2 + cos(π)**2 = 1
=> sin(π)**2 + -1**2     = 1
=> sin(π)**2             = 0
=> sin(π)                = 0

Consequently, we can evaluate e(iπ) as:

 e(iπ) = cos(π) + i * sin(π) = -1 + i * 0 = -1;

Similarily, we can evaluate e(-iθ) from e(iθ):

   e( iθ) = cos( θ) + i * sin( θ)
=> e(-iθ) = cos(-θ) + i * sin(-θ)
=> e(-iθ) = cos(θ)  - i * sin( θ)  as cos() is even, sin() is odd

This result allows us to evaluate e(iθ)e(-iθ):

   e(iθ)e(-iθ) = (cos(θ) + i * sin(θ))(cos(θ) - i * sin(θ))
               = (cos(θ)**2 - i*i*sin(θ)**2)
               =  cos(θ)**2 +     sin(θ)**2         as i*i = -1
               =  1                                 Pythagoras

Starting again from the "Cosine Double Angle Formula", we know that:

cos(2θ) = 2 * cos(θ)**2 - 1

Replace cos() by its formulation in e():

cos(θ) = e(iθ)/2 + e(-iθ)/2 

to get:

e(2iθ)/2 + e(-2iθ)/2 = 2 * (e(iθ)/2 + e(-iθ)/2)**2                     - 1  
                     = 2 * (e(iθ)/2)**2+(e(-iθ)/2)**2 +2e(iθ)e(-iθ)/4) - 1
                     =     (e(iθ)**2)/2+(e(-iθ)**2)/2 + e(iθ)e(-iθ)    - 1

But

e(iθ)e(-iθ) = 1

so we continue the algebraic simplification to get:

 e(2iθ)/2 + e(-2iθ)/2 =     (e(iθ)**2)/2+(e(-iθ)**2)/2 + e(iθ)e(-iθ)    - 1
                      =     (e(iθ)**2)/2+(e(-iθ)**2)/2 + 1              - 1
                      =     (e(iθ)**2)/2+(e(-iθ)**2)/2 

Again

e(iθ)e(-iθ) = 1

so we are forced to conclude that

 e( 2iθ) = e( iθ)**2  and
 e(-2iθ) = e(-iθ)**2   for any angle θ.

The e() function is behaving like exponentiation, that is we can write:

 e(iθ) = e**iθ

where e is some number whose value is as yet unknown, which is the solution to the equation:

 e**iπ = -1

As the e() function behaves like an exponential:

e(i*θ1) * e(i*θ2) = e**(i*θ1) * e** (i*θ2) = e**(i*(θ1+θ2)) = e(i(θ1+θ2))

In particular:

 (cos(θ) + i * sin(θ))**n =  e(iθ)**n 
                          = (e**iθ)**n
                          =  e**inθ 
                          =  e(inθ)
                          = (cos(nθ) + i * sin(nθ))

Lets try this formula out with n = 2:

  (cos(θ) + i * sin(θ))**2                       = cos(2θ) + i * sin(2θ)

=> (cos(θ)**2 - sin(θ))**2 + 2 * i * cos(θ)sin(θ) = cos(2θ) + i * sin(2θ)

Now, the number i is manifestly not a real number, as no real number is a solution to the equation i*i = -1, yet both the cos() and sin() functions produce real numbered results, they are, after all, just the ratios of the lengths of the sides of triangles. Consequently in the above we can equate the parts of the equations which are separated by being multiplied by i to get two equations:

  cos(2θ) = cos(θ)**2 - sin(θ))**2
  sin(2θ) = 2*cos(θ)*sin(θ)

Recall the "Cosine Angle Sum Formula" of:

  cos(θ1+θ2) = cos(θ1)cos(θ2) - sin(θ1)sin(θ2)

Set θ = θ1 = θ2 to get the identical result:

  cos(θ+θ) = cos(2θ) = cos(θ)cos(θ) - sin(θ)sin(θ)

Likewise the Recall the "Sine Angle Sum Formula" of:

  sin(θ1+θ2) = cos(θ1)sin(θ2) + sin(θ1)cos(θ2)

Set θ = θ1 = θ2 to get the identical result:

  sin(θ+θ) = sin(2θ) = cos(θ)sin(θ) + sin(θ)cos(θ) = 2*sin(θ)cos(θ)

Using Cosine and Sine Angle Sum Formulae and equating parts, we can deduce that:

  e(i*θ1) * e(i*θ2))
   = (cos(θ1) + i * sin(θ1)) * (cos(θ2) + i * sin(θ2))
   = (cos(θ1)*cos(θ2) - sin(θ1)*sin(θ2) + i(cos(θ1)sin(θ2)) + cos(θ2)sin(θ1))
   =  cos(θ1 + θ2) + isin(θ1 + θ2)   
   =  e(i(θ1+θ2))

wherein the e() function demonstrates its exponential nature to perfection.

The ability of i to partition single equations into two orthogonal simultaneous equations makes expressions of the form e(iθ), and hence trigonometery, invaluable in such diverse applications as electronics: simultaneously representing current and voltage in the same equation; and quantum mechanics, where it is necessary to represent position and momentum, or time and energy as pairs of variables partitioned by the uncertainty principle.

[edit] Taylor series approximations for the trig. functions

Having learnt a good deal about the properties of the cos() function, it might be helpful to know how to calculate cos(θ) for a given θ. We know that cos() can be defined in terms of the e() function which has the remarkable property that d e(θ) / dθ = e(θ), and e(0) = 1. Perhaps it is possible to represent e() by an infinite polynomial:

e(θ) = a0 + a1*θ + a2*θ**2 + a3*θ**3 + ...  an*θ**n +  ....

in which case evaluating at θ = 0 yields a0 = 1. To find the rate of change with θ of the n'th term:

d an*θ**n / dθ = (an*(θ+δθ)**n - an*(θ)**n)           / δθ as δθ tends to zero
               = an *((θ+δθ)**n - θ**n)               / δθ as δθ tends to zero
               = an *((θ**n+n*δθ*θ**(n-1)+... - θ**n) / δθ as δθ tends to zero
               = an *(n*δθ*θ**(n-1)+...)              / δθ as δθ tends to zero
               = an * n*θ*(n-1)+...

where the ... represent terms multiplied by at least δθ**2, and which therefore tend to zero as δθ tends to zero.

Finding the rate of change with respect to θ of the polynomial representing e(θ) produces:

e(θ) = a1*θ + 2 * a2*θ**1 + 3 * a3*θ**2 + ...  n * an*θ**(n-1) +  ....

Again evaluating at θ = 0, we get a1 = 1

Repeating the whole process, we get:

e(θ) = 2 * a2 + 3 * 2* a3*θ + ...  n * (n-1) * an*θ**(n-2) +  ....

and evaluating at θ = 0, we get a2 = 1/2, and in general, an = 1/n!, wher n! means n*(n-1)*(n-2)*...3*2*1.

Putting these results together, we get:

e(θ) = 1 + θ + θ**2/(2*1) + θ**3/(3*2*1)2 + ... θ**n / n! +  ....

Recalling that e(θ) = e**θ, and thus that e(1) = e**1 = e, we find e to be:

e = 1+1+1/2+1/6+1/24+.... 1/n!+ ...

which is approximately equal to 2.71828183

To calculate cos(θ), we would apply the formula:

   2*cos(θ) = e(iθ)+e(-iθ)
            = 1 + iθ - θ**2/2 - iθ**3/6 + θ**4/24 + ...   
            + 1 +-iθ - θ**2/2 --iθ**3/6 + θ**4/24 + ...
            = 2      -2θ**2/2           +2θ**4/24 + ...
 => cos(θ)  = 1 - θ**2/2 + θ**4/24 + ... (-1)**n*θ**(2n)/((2n)!)+ ...

in which the odd terms cancel out and the even terms double and alternate in sign. Substituting -θ for θ does not change the result as θ occurs with even powers, hence the earlier phrase that cos() is an even function.

[edit] Using the Taylor series to calculate π

Having calculated, e and cos(), lets see if we can calculate π. Draw a circle with radius of length one, and draw diameters to divide the circle up into n segments of equal area, and hence equal angles θ; connect the ends of each diameter to its neighbors to get a regular polygon of n equal length sides. Each segment of the polygon is an isoceles triangle imbedded within a segment of the circle. The length of the sides og the polygon is less than the distance along the circumference of the circle between each corner of the polygon, the circumference length is exactly θ because we are using radians to measure angles within a circle of radius of length one. We can bisect each side of the polygon to construct a regular polygon with 2*n equal sides,the angle between each diameter is now θ/2 - lets us call the angle between diameters the interior angle. As the interior angle gets smaller and smaller, the lengths measured between two adjacent corners of the polygon, measured first along the side of the polygon, the choord length, and secondly along the circumference of the circle, the arc length, agree more and more closely. We can get this agreement as close as we might like by dividing the interior angle often enough, although to get exact agreement we would have to divide the interior angle an infinite number of times, clearly impossible in the real world of physics, but within the realm of mathematics it is possible to contemplate such an infinite process.

We have already defined the length of the circumference of a circle of diameter of length one as having the value π, so now all we need to do is find the total length of the sides of the polygons we are using to approximate π and then extrapolate as the number of sides approaches infinity to get an estimate for the value of π.

Let us start the process off using a polygon of 4 sides, a square, which has an interior angle of a right angle, that is, π/2. Draw a square, then draw its diagonals, and a circle through the corners of the square, to get 4 chords and 4 arcs. Scale the drawing so that the radius of the circle drawn is exactly 1/2. Each chord is the base of isoceles triangle which can be bisected into two identical right angled triangles. The hypotenuse or side c of of each of these right angle triangles is a radius, we have scaled the drawing above so that each such radius will be of length 1/2. The angle θ of each such right angle triangle is 1/2*π/2 = π/4. The length of side b in each such right angle triangle has length 1/2*sin(π/4) = 1/2*sin(2π/8). If we bisect each interior angle of the square we get a total of 8 of these right angled triangles, hence our first approximation to the length of the circumference of a circle of diameter of length one which is defined to be π, is 8 * the length of side b:

    π = 8 * 1/2 * sin(2*π/8)
=> 2π = n * sin(2*π/n) where n = 8.

If we extend each side a of each of the each such right angle triangles to meet the circumference of the circle at points, and then draw lines from each of these points to the points neighboring points where the diagonals of the square cross the circumference we will construct the octogon - a polygon with 8 equal sides - twice the number of the square, with interior angles of π/4, half that of the square. The argument used above to calculate the lengths of the sides of the square can be applied to the octogon. When this argument was first applied to the square it seems unnecessarily complicated: it would have been easier to notice that diagonals of the square conveniently meet at right angles allowing us to apply Pythagoras immediately to calculate the length of the side of a square drawn in a circle of radius of length 1/2 as squareRoot(1/2**2+1/2**2) = squareRoot(1/4+1/4) = squareRoot(1/2). While this answer is perfectly correct for a square, it does not work for any other polygon as their diagonals do not meet at such a convenient angle, while the formula:

=> 2π = n * sin(2*π/n) 

works for any value of n.

All we need now is the value of the sin() function for polygons with many sides, that is with very small interior angles, or more specifically, with n large. We do know the sin() function for one angle:

sin(π/2) = sin(2π/4) = 1

The "Sine Half Angle Formula" lets us calculate the value of the sin() function for half angle of known angles:

    sin(θ/2)      = squareRoot(1/2)*squareRoot(1 - squareRoot(1 - sin(θ)**2))  
  
 => sin(2π/8)     = squareRoot(1/2)*squareRoot(1 - squareRoot(1 - 1**2))      
                  = squareRoot(1/2)      
 => sin(2π/16)    = squareRoot(1/2)*squareRoot(1 - squareRoot(1 - squareRoot(1/2)**2))      
                  = squareRoot(1/2)*squareRoot(1 - squareRoot(1 - 1/2))      
                  = squareRoot(1/2 - 1/2*squareRoot(1/2))      
 => sin(2π/32)    = squareRoot(1/2)*squareRoot(1 - squareRoot(1 - (squareRoot(1/2 - 1/2*squareRoot(1/2)))**2))  
                  = squareRoot(1/2)*squareRoot(1 - squareRoot(1 - 1/2 + 1/2*squareRoot(1/2)))  
                  = squareRoot(1/2)*squareRoot(1 - squareRoot(1/2 + 1/2*squareRoot(1/2)))  
                  = squareRoot(1/2 - 1/2*squareRoot(1/2 + 1/2*squareRoot(1/2)))  

Its easy to guess from this pattern that:

 => sin(2π/64)    = squareRoot(1/2 - 1/2*squareRoot(1/2 + 1/2*squareRoot(1/2 + 1/2*squareRoot(1/2))))

and so on. This formula does give an approximate value for π, an 8 digit calculator produced a value of 3.13 which is within 1% of the true value. However, this formula cannot be regarded as an efficient way of calculating π accurately to many digits: the constant need to take square roots insures a continual loss of precision.

We have now built up some basic trigonometric results.

[edit] Trigonometric definitions

We have defined the sine, cosine, and tangent functions using the unit circle. Now we can apply them to a right triangle.

right triangle

This triangle has sides a and b. The angle between them, C, is a right angle. The third side, c, is the hypotenuse. Side a is opposite angle A, and side b is adjacent to angle A.

Applying the definitions of the functions, we arrive at these useful formulas:

   sin(A) = a / c  or opposite side over hypotenuse
   cos(A) = b / c  or adjacent side over hypotenuse
   tan(A) = a / b  or opposite side over adjacent side 

This can be memorized using the mnemonic 'SOHCAHTOA' (sin = opposite over hypothenuse, cosine = adjacent over hypotenuse, tangent = opposite over adjacent) (also denoted SohCahToa).

Exercises: (Draw a diagram!)

1. A right triangle has side a = 3, b = 4, and c = 5. Calculate the following:

sin(A), cos(A), tan(A)

2. A different right triangle has side c = 6 and sin(A) = 0.5 . Calculate side a.


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[edit] Graphs of Sine and Cosine Functions

The graph of the sine function looks like this:

Graph of the sine function

Careful analysis of this graph will show that the graph corresponds to the unit circle. X is essentially the degree measure (in radians), while Y is the value of the sine function.

The graph of the cosine function looks like this:

Graph of the cosine function

As with the sine function, analysis of the cosine function will show that the graph corresponds to the unit circle. One of the most important differences between the sine and cosine functions is that sine is an odd function while cosine is an even function.

Sine and cosine are periodic functions; that is, the above is repeated for preceding and following intervals with length 2π.


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[edit] Graphs of Other Trigonometric Functions

A graph of tan(x). tan(x) is defined as \frac{\sin(x)}{\cos(x)}.

Graph of tan(x)

A graph of csc(x). csc(x) is defined as \frac{1}{\sin(x)}.

Graph of csc(x)

A graph of sec(x). sec(x) is defined as \frac{1}{\cos(x)}.

Graph of sec(x)

A graph of cot(x). cot(x) is defined as \frac{1}{\tan(x)} or \frac{\cos(x)}{\sin(x)}.

Graph of cot(x)

Note that tan(x), sec(x), and csc(x) are unbounded.

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[edit] Inverse Trigonometric Functions

< Trigonometry

[edit] The Inverse Functions, Restrictions, and Notation

While it might seem that inverse trigonometric functions should be relatively self defining, some caution is necessary to get an inverse function since the trigonometric functions are not one-to-one. To deal with this issue, some texts have adopted the convention of allowing sin − 1x, cos − 1x, and tan − 1x (all with lower-case initial letters) to indicate the inverse relations for the trigonometric functions and defining new functions Sinx, Cosx, and Tanx (all with initial capitals) to equal the original functions but with restricted domain, thus creating one-to-one functions with the inverses Sin − 1x, Cos − 1x, and Tan − 1x. For clarity, we will use this convention. Another common notation used for the inverse functions is the "arcfunction" notation: Sin − 1x = arcsinx, Cos − 1x = arccosx, and Tan − 1x = arctanx (the arcfunctions are sometimes also capitalized to distinguish the inverse functions from the inverse relations). The arcfunctions may be so named because of the relationship between radian measure of angles and arclength--the arcfunctions yield arc lengths on a unit circle.

The restrictions necessary to allow the inverses to be functions are standard: Sin − 1x has range \left[-\begin{matrix} \frac{\pi}{2} \end{matrix},\begin{matrix} \frac{\pi}{2} \end{matrix}\right]; Cos − 1x has range \left[0,\pi\right]; and Tan − 1x has range \left(-\begin{matrix} \frac{\pi}{2} \end{matrix},\begin{matrix} \frac{\pi}{2} \end{matrix}\right) (these restricted ranges for the inverses are the restricted domains of the capital-letter trigonometric functions). For each inverse function, the restricted range includes first-quadrant angles as well as an adjacent quadrant that completes the domain of the inverse function and maintains the range as a single interval.

It is important to note that because of the restricted ranges, the inverse trigonometric functions do not necessarily behave as one might expect an inverse function to behave. While \mbox{Sin}^{-1}\left(\sin\left(\begin{matrix} \frac{\pi}{6} \end{matrix}\right)\right)=\mbox{Sin}^{-1}\left(\begin{matrix} \frac{1}{2} \end{matrix}\right)=\begin{matrix} \frac{\pi}{6} \end{matrix} (following the expected \mbox{Sin}^{-1}\left(\sin x\right)=x), \mbox{Sin}^{-1}\left(\sin\left(\begin{matrix} \frac{5\pi}{6} \end{matrix}\right)\right)=\mbox{Sin}^{-1}\left(\begin{matrix} \frac{1}{2} \end{matrix}\right)=\begin{matrix} \frac{\pi}{6} \end{matrix}! For the inverse trigonometric functions, f^{-1}\circ f(x)=x only when x is in the range of the inverse function. The other direction, however, is less tricky: f\circ f^{-1}(x)=x for all x to which we can apply the inverse function.

[edit] The Inverse Relations

For the sake of completeness, here are definitions of the inverse trigonometric relations based on the inverse trigonometric functions:

  • \sin^{-1} x = \left\{ \mbox{Sin}^{-1} x + 2\pi n, n \in \mathbb{Z} \right\} \cup \left\{ \pi - \mbox{Sin}^{-1} x + 2\pi n, n \in \mathbb{Z} \right\} (the sine function has period , but within any given period may have two solutions and \sin x=\sin\left(\pi-x\right))
  • \cos^{-1} x = \left\{ \pm \mbox{Cos}^{-1} x + 2\pi n, n \in \mathbb{Z} \right\} (the cosine function has period , but within any given period may have two solutions and cosine is even--\cos x=\cos\left(-x\right))
  • \tan^{-1} x = \left\{ \mbox{Tan}^{-1} x + \pi n, n \in \mathbb{Z} \right\} (the tangent function has period π and is one-to-one within any given period)


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[edit] Applications and Models

< Trigonometry

[edit] Simple harmonic motion

Simple harmonic motion. Notice that the position of the dot matches that of the sine wave.

Simple harmonic motion (SHM) is the motion of an object which can be modeled by the following function:

x = A \sin \left(\omega t + \phi\right)

or

x = c_{1} \cos\left(\omega t\right) + c_{2} \sin\left(\omega t\right)
where c1 = A sin φ and c2 = A cos φ.

In the above functions, A is the amplitude of the motion, ω is the angular velocity, and φ is the phase.

The velocity of an object in SHM is

v = A \omega \cos \left(\omega t + \phi\right)

The acceleration is

a = -A \omega^{2} \sin \left(\omega t + \phi\right)

[edit] Springs and Hooke's Law

An application of this is the motion of a weight hanging on a spring. The motion of a spring can be modeled approximately by Hooke's law:

F = -kx

where F is the force the spring exerts, x is the position of the end of the spring, and k is a constant characterizing the spring (the stronger the spring, the higher the constant).

[edit] Calculus-based derivation

From Newton's laws we know that F = ma where m is the mass of the weight, and a is its acceleration. Substituting this into Hooke's Law, we get

ma = -kx

Dividing through by m:

a = -\frac{k}{m}x

The calculus definition of acceleration gives us

x'' = -\frac{k}{m}x
x'' + \frac{k}{m}x = 0

Thus we have a second-order differential equation. Solving it gives us

x = c_{1} \cos\left(\sqrt{\frac{k}{m}}t\right) + c_{2} \sin\left(\sqrt{\frac{k}{m}}t\right) (2)

with an independent variable t for time.


We can change this equation into a simpler form. By lettting c1 and c2 be the legs of a right triangle, with angle φ adjacent to c2, we get

\sin \phi = \frac{c_{1}}{\sqrt{c_{1}^{2} + c_{2}^{2}}}
\cos \phi = \frac{c_{2}}{\sqrt{c_{1}^{2} + c_{2}^{2}}}

and

c_{1} = \sqrt{c_{1}^{2} + c_{2}^{2}} \sin \phi
c_{2} = \sqrt{c_{1}^{2} + c_{2}^{2}} \cos \phi

Substituting into (2), we get

x = \sqrt{c_{1}^{2} + c_{2}^{2}} \sin \phi \cos\left(\sqrt{\frac{k}{m}}t\right) + \sqrt{c_{1}^{2} + c_{2}^{2}} \cos \phi \sin\left(\sqrt{\frac{k}{m}}t\right)

Using a trigonometric identity, we get:

x = \sqrt{c_{1}^{2} + c_{2}^{2}} \left[\sin \left(\phi + \sqrt{\frac{k}{m}}t\right) + \sin \left(\phi - \sqrt{\frac{k}{m}}t\right)\right] + \sqrt{c_{1}^{2} + c_{2}^{2}} \left[\sin \left(\sqrt{\frac{k}{m}}t + \phi\right) + \sin \left(\sqrt{\frac{k}{m}}t - \phi\right)\right]
x = \sqrt{c_{1}^{2} + c_{2}^{2}} \sin \left(\sqrt{\frac{k}{m}}t + \phi\right) (3)

Let A = \sqrt{c_{1}^{2} + c_{2}^{2}} and \omega^{2} = \frac{k}{m}. Substituting this into (3) gives

x = A \sin \left(\omega t + \phi\right)


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[edit] Analytic Trigonometry

[edit] Using Fundamental Identities

Some of the fundamental trigometric identities are those derived from the Pythagorean Theorem. These are defined using a right triangle:

right triangle

By the Pythagorean Theorem,

A2 + B2 = C2 {1}

Dividing through by C2 gives

\left(\frac{A}{C}\right)^2+\left(\frac{B}{C}\right)^2=\left(\frac{C}{C}\right)^2=1 {2}

We have already defined the sine of a in this case as A/C and the cosine of a as B/C. Thus we can substitute these into {2} to get

\sin^2 a + \cos^2 a \equiv 1

Related identities include:

\sin^2 a \equiv 1 - \cos^2 a\mbox{  or  }\cos^2 a \equiv 1-\sin^2 a
\tan^2 a + 1 \equiv \sec^2 a\mbox{  or  }\tan^2 a \equiv \sec^2 a -1
1 + \cot^2 a \equiv \csc^2 a\mbox{  or  }\cot^2 a \equiv \csc^2 a -1

Other Fundamental Identities include the Reciprocal, Ratio, and Co-function identities

Reciprocal identities

\csc a \equiv \frac{1}{\sin a}\quad\sec a \equiv \frac{1}{\cos a}\quad\cot a \equiv \frac{1}{\tan a}

Ratio identities

\tan a \equiv \frac{\sin a}{\cos a}\quad\cot a \equiv \frac{\cos a}{\sin a}

Co-function identities (in radians)

\cos a \equiv \sin\left(\frac{\pi}{2}-a\right)\quad \csc a \equiv \sec\left(\frac{\pi}{2}-a\right)\quad \cot a \equiv \tan\left(\frac{\pi}{2}-a\right)

[edit] Verifying Trigonometric Identities

To verify an identity means to make sure that the equation is true by setting both sides equal to one another.

There is no set method that can be applied to verifying identities; there are, however, a few different ways to start based on the identity which is to be verified.

[edit] Introduction

Trigonometric identities are used in both course texts and in real life applications to abbreviate trigonometric expressions. It is important to remember that merely verifying an identity or altering an expression is not an end in itself, but rather that identities are used to simplify expressions according to the task at hand. Trigonometric expressions can always be reduced to sines and cosines, which can be more manageable than their inverse counterparts.

[edit] To verify an identity:

1) Always try to reduce the larger side first.

2) Sometimes getting all trigonometric functions on one side can help.

3) Remember to use and manipulate already existing identities. The Pythagorean identities are usually the most useful in simplifying.

4) Remember to factor if needed.

5) Whenever you have a squared trigonometric function such as Sin2(t), always go to your Pythagorean identities, which deal with squared functions.

6) Sometimes doing the reverse of the normal steps helps. For example, adding 1 in unique forms (such as cos(t) / cos(t) can help simplify expressions by matching denominators and simplifying numerators.

Easy example 1 / cot(t) = sin(t) / cos(t)

1/cot(t)= tan(t) ,so tan(t) = sin(t) / cos(t)

tan(t) is the same as sint(t)/cos(t) and therefore can be rewritten as sin(t) / cos(t) = sin(t) / cos(t)

Identity verified

[edit] Solving Trigonometric Equations

Trigonometric equations involve finding an unknown which is an argument to a trigonometric function.

[edit] Basic trigonometric equations

[edit] sin x = n

Sin unit circle.png
n\,\! \sin x=n\,\!
\left|n\right|<1 \begin{matrix}x=\alpha + 2 k \pi \\
x=\pi - \alpha + 2 k \pi \\
\alpha \in \left[-\frac{\pi}{2};\frac{\pi}{2}\right]\end{matrix}
n=-1\,\! x=-\begin{matrix}\frac{\pi}{2}\end{matrix}+2k\pi
n=0\,\! x=k\pi\,\!
n=1\,\! x=\begin{matrix}\frac{\pi}{2}\end{matrix}+2k\pi
\left|n\right|>1 x\in\varnothing

The equation sinx = n has solutions only when n is within the interval [-1; 1]. If n is within this interval, then we need to find an α such that:

\alpha=\sin^{-1} n\,\!

The solutions are then:

x=\alpha + 2 k \pi\,\!
x=\pi - \alpha + 2 k \pi\,\!

Where k is an integer.

In the cases when n equals 1, 0 or -1 these solutions have simpler forms which are summarizied in the table on the right.

For example, to solve:

\sin \frac{x}{2}=\frac{\sqrt{3}}{2}

First find α:

\alpha=\sin^{-1}{\frac{\sqrt{3}}{2}}=\frac{\pi}{3}

Then substitute in the formulae above:

\frac{x}{2}=\frac{\pi}{3} + 2 k \pi
\frac{x}{2}=\pi - \frac{\pi}{3} + 2 k \pi

Solving these linear equations for x gives the final answer:

x=\frac{2\pi}{3}\left(1+6k\right)
x=\frac{4\pi}{3}\left(1+3k\right)

Where k is an integer.

[edit] cos x = n

Cos unit circle.png
n\,\! \cos x=n\,\!
\left|n\right|<1 \begin{matrix}x=\pm\alpha + 2 k \pi \\
\alpha \in \left[0;\pi\right]\end{matrix}
n=-1\,\! x=\pi+2k\pi\,\!
n=0\,\! x=\begin{matrix}\frac{\pi}{2}\end{matrix}+k\pi\,\!
n=1\,\! x=2k\pi\,\!
\left|n\right|>1 x\in\varnothing

Like the sine equation, an equation of the form cosx = n only has solutions when n is in the interval [-1; 1]. To solve such an equation we first find the angle α such that:

\alpha=\cos^{-1}n\,\!

Then the solutions for x are:

x=\pm\alpha+2k\pi \, \!

Where k is an integer.

Simpler cases with n equal to 1, 0 or -1 are summarized in the table on the right.

[edit] tan x = n

Tan unit circle.png
n\,\! \tan x=n\,\!
General
case
\begin{matrix}x=\alpha + k\pi \\
\alpha \in \left[-\frac{\pi}{2};\frac{\pi}{2}\right]\end{matrix}
n=-1\,\! x=-\begin{matrix}\frac{\pi}{4}\end{matrix}+k\pi\,\!
n=0\,\! x=k\pi\,\!
n=1\,\! x=\begin{matrix}\frac{\pi}{4}\end{matrix}+k\pi\,\!

An equation of the form tanx = n has solutions for any real n. To find them we must first find an angle α such that:

\alpha=\tan^{-1}n\,\!

After finding α, the solutions for x are:

x=\alpha+k\pi\,\!

When n equals 1, 0 or -1 the solutions have simpler forms which are shown in the table on the right.

[edit] cot x = n

Cot unit circle.png
n\,\! \cot x=n\,\!
General
case
\begin{matrix}x=\alpha + k\pi \\
\alpha \in \left[0;\pi\right]\end{matrix}
n=-1\,\! x=-\begin{matrix}\frac{3\pi}{4}\end{matrix}+k\pi\,\!
n=0\,\! x=\begin{matrix}\frac{\pi}{2}\end{matrix}+k\pi\,\!
n=1\,\! x=\begin{matrix}\frac{\pi}{4}\end{matrix}+k\pi\,\!

The equation cotx = n has solutions for any real n. To find them we must first find an angle α such that:

\alpha=\cot^{-1}n\,\!

After finding α, the solutions for x are:

x=\alpha+k\pi\,\!

When n equals 1, 0 or -1 the solutions have simpler forms which are shown in the table on the right.

[edit] csc x = n and sec x = n

The trigonometric equations csc x = n and sec x = n can be solved by transforming them to other basic equations:

\csc x=n\Leftrightarrow\frac{1}{\sin x}=n\Leftrightarrow\sin x=\frac{1}{n}
\sec x=n\Leftrightarrow\frac{1}{\cos x}=n\Leftrightarrow\cos x=\frac{1}{n}

[edit] Further examples

Generally, to solve trigonometric equations we must first transform them to a basic trigonometric equation using the trigonometric identities. This sections lists some common examples.

[edit] a sin x + b cos x = c

To solve this equation we will use the identity:

a \sin x + b \cos x = \sqrt{a^2 + b^2} \sin \left(x + \alpha\right)
\alpha = \begin{cases} \tan^{-1} \left(b / a\right), & \mbox{if } a > 0 \\ \pi + \tan^{-1} \left(b / a\right), & \mbox{if } a < 0 \end{cases}

The equation becomes:

\sqrt{a^2 + b^2} \sin \left(x + \alpha\right) = c
\sin{\left(x + \alpha\right)} = \frac{c}{\sqrt{a^2 + b^2}}

This equation is of the form sinx = n and can be solved with the formulae given above.

For example we will solve:

\sin 3 x - \sqrt{3}\cos 3 x = -\sqrt{3}

In this case we have:

a = 1, b = -\sqrt{3}
\sqrt{a^2+b^2} = \sqrt{1^2 + \left(-\sqrt{3}\right)^2} = 2
\alpha = \tan^{-1} \left(-\sqrt{3}\right) = -\frac{\pi}{3}

Apply the identity:

2 \sin \left(3x - \frac{\pi}{3}\right) = -\sqrt{3}
\sin \left(3x - \frac{\pi}{3}\right) = -\frac{\sqrt{3}}{2}

So using the formulae for sinx = n the solutions to the equation are:

3x - \frac{\pi}{3} = -\frac{\pi}{3} + 2 k \pi \Leftrightarrow x = \frac{2 k \pi}{3}
3x - \frac{\pi}{3} = \pi + \frac{\pi}{3} + 2 k \pi \Leftrightarrow x = \frac{\pi}{9}\left(6k + 5\right)

Where k is an integer.

[edit] Sum and Difference Formulas

[edit] Cosine Formulas

\cos (A + B) = \cos A \cos B - \sin A \sin B \,
\cos (A - B) = \cos A \cos B + \sin A \sin B \,
\cos 2A = \cos^2 A - \sin^2 A = 2 \cos^2 A -1 = 1-2 \sin^2 A \,
\cos \frac{A}{2} = \pm \sqrt{\frac{1+\cos A}{2}} \,

[edit] Sine Formulas

\sin(a + b) = \sin a \cos b + \cos a \sin b\,

\sin(a - b) = \sin a \cos b - \cos a \sin b\,

\sin 2a = 2 \sin a \cos a\,

\sin \frac{A}{2} = \pm \sqrt{\frac{1-\cos A}{2}} \,

[edit] Tangent Formulas

\tan(a+b)= \frac{\tan(a) + \tan(b)}{1-\tan(a)\tan(b)}
\tan(a-b)= \frac{\tan(a) - \tan(b)}{1+\tan(a)\tan(b)}
\tan 2A = {2 \tan A \over 1 - \tan^2 A} = {2 \cot A \over \cot^2 A - 1} = {2 \over \cot A - \tan A} \,
\tan \frac{A}{2} = \pm \sqrt{\frac{1-\cos A}{1+\cos A}} = \frac {\sin A}{1+\cos A} = \frac {1-\cos A}{\sin A} \,

[edit] Derivations

  • cos(a + b) = cos a cos b - sin a sin b
  • cos(a - b) = cos a cos b + sin a sin b

Using cos(a + b) and the fact that cosine is even and sine is odd, we have

            cos(a + (-b)) = cos a cos (-b) - sin a sin (-b)
                          = cos a cos b - sin a (-sin b)
                          = cos a cos b + sin a sin b
  • sin(a + b) = sin a cos b + cos a sin b

Using cofunctions we know that sin a = cos (90 - a). Use the formula for cos(a - b) and cofunctions we can write

         sin(a + b) = cos(90 - (a + b))
                    = cos((90 - a) - b)
                    = cos(90 -a)cos b + sin(90 - a)sin b
                    = sin a cos b + cos a sin b
  • sin(a - b) = sin a cos b - cos a sin b

Having derived sin(a + b) we replace b with "-b" and use the fact that cosine is even and sine is odd.

      sin(a + (-b)) = sin a cos (-b) + cos a sin (-b) 
                    = sin a cos b + cos a (-sin b)
                    = sin a cos b - cos a sin b

Multiple-Angle and Product-to-sum Formulas


[edit] Proofs for Double Angle Formulas

[edit] Using the Sum and Difference Identities

Recall that:
 \displaystyle\cos(a + b) = \cos(a)\cos(b) - \sin(a)\sin(b)

Using a = b in the above formula yields:

   \displaystyle\cos(2a) = \cos(a + a) = \cos(a)\cos(a) - \sin(a)\sin(a) = \cos^2(a) - \sin^2(a) 

From the last (rightmost) term, two more identities may be derived. One containing only a sine:

   \displaystyle\cos(2a) = \cos^2(a) - \sin^2(a) = \cos^2(a) - \sin^2(a) + 0 = \cos^2(a) - \sin^2(a) + [\sin^2(a) - \sin^2(a)] = [\cos^2(a) + \sin^2(a)] - 2\sin^2(a) = 1 - 2\sin^2(a) 

And one containing only a cosine:

   \displaystyle\cos(2a) = \cos^2(a) - \sin^2(a) = \cos^2(a) - \sin^2(a) + 0 = \cos^2(a) - \sin^2(a) + [\cos^2(a) - \cos^2(a)] = 2\cos^2(a) - [\cos^2(a) + \sin^2(a)] = 2\cos^2(a) - 1 


\displaystyle\sin(a + b) = \sin(a)\cos(b) + \sin(b)\cos(a) ; a = b for sin(2a)

  \displaystyle\sin(a)\cos(a) + \sin(a)\cos(a) = 2\cos(a)\sin(a) = \sin(2a)

Next Page: Law of Sines | Previous Page: Sum and Difference Formulas

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[edit] Additional Topics in Trigonometry

[edit] Law of Sines

Consider this triangle:

Triangle with Perpendicular.png

It has three sides

  • A, length A, opposite angle a at vertex a
  • B, length B, opposite angle b at vertex b
  • C, length C, opposite angle c at vertex c

The perpendicular, oc, from line ab to vertex c has length h

The Law of Sines states that:

{A \over \sin a} = {B \over \sin b} = {C \over \sin c}

The law can also be written as the reciprocal:

{\sin a \over A} = {\sin b \over B} = {\sin c \over C}

[edit] Proof

The perpendicular, oc, splits this triangle into two right-angled triangles. This lets us calculate h in two different ways

  • Using the triangle cao gives
h=B \sin a \,
  • Using the triangle cbo gives
h=A \sin b \,
  • Eliminate h from these two equations
A \sin b =B \sin a \,
  • Rearrange
{A \over \sin a} = {B \over \sin b}

By using the other two perpendiculars the full law of sines can be proved. QED.

[edit] Law of Cosines

Consider this triangle:

Triangle with Perpendicular.png

It has three sides

  • A, length A, opposite angle a at vertex a
  • B, length B, opposite angle b at vertex b
  • C, length C, opposite angle c at vertex c

The perpendicular, oc, from line ab to vertex c has length h

The Law of Cosines states that:

A^2=B^2 + C^2 - 2BC \cdot\cos a
B^2=A^2 + C^2 - 2AC \cdot\cos b
C^2=A^2 + B^2 - 2AB \cdot\cos c

[edit] Proof

The perpendicular, oc, divides this triangle into two right angled triangles, aco and bco.

First we will find the length of the other two sides of triangle aco in terms of known quantities, using triangle bco.

h=A sin b

Side C is split into two segments, total length C.

ob, length A cos b
ao, length C - A cos b

Now we can use Pythagoras to find B, since B2 = ao2 + h2

\begin{matrix}B^2 & = & (C-A\cos b)^2+A^2\sin^2 b \\
\ &=& C^2-2AC\cos b +A^2\cos^2 b+A^2\sin^2 b\\
\ &=& A^2+C^2-2AC\cos b\end{matrix}

The corresponding expressions for A and C can be proved similarly.

Next Page: Solving Triangles | Previous Page: Law of Cosines

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[edit] Solving Triangles

One of the most common applications of trigonometry is solving triangles—finding missing sides and/or angles given some information about a triangle. The process of solving triangles can be broken down into a number of cases.

[edit] Given a Right Triangle

If we know the triangle is a right triangle, one side and one of the non-right angles or two sides uniquely determine the triangle. (It is also worth noting that if we are given a triangle not specified to be a right triangle, but two given angles are complementary, then the third angle must be a right angle, so we have a right triangle.) Use right triangle trigonometry to find any missing information.

[edit] Given Three Angles (AAA) and No Sides

From two angles the third can be computed. Hence, given two angles one is given all three angles (since the sum of the measures of the internal angles of a triangle form a straight angle). Any two triangles with congruent corresponding angles are themselves congruent triangles. Congruent triangles are similar triangles, meaning that they possess precisely the same shape. However, without a length to extrapolate a relative size, no computation of sides, length or size can be performed.

[edit] Given Three Sides (SSS)

Given three sides of a triangle, we can use the Law of Cosines to find any of the missing angles (the alternate form \cos A = \frac{b^2+c^2-a^2}{2bc} may be helpful here). (The a2 is the side opposite the angle you are to find)

[edit] Given Two Sides and the Included Angle (SAS)

Given two sides and the angle included by the two given sides, we can apply the Law of Cosines to find the missing side, then procede as above to find the missing angles.

[edit] Given Two Angles and a Side (AAS or ASA)

Given two angles, we can find the third angle (since the sum of the measures of the three angles in a triangle is a straight angle). Knowing all three angles and one side, we can use the Law of Sines to find the missing sides. But how do we find out how to work the angle. No one knows. I ment 2 put that on the one above this.

[edit] Given Two Sides and an Angle Not Included by the Two Sides (SSA or the Ambiguous Case)

Here, we run into trouble—the given information may not uniquely determine a triangle. One of the two given sides is opposite the given angle, so we can apply the Law of Sines to try to find the angle opposite the other given side. When we do this (given sides a and b and angle A), we'll have \sin B = \frac{b\sin A}{a} and there are three possibilities: sinB > 1, sinB = 1, or sinB < 1.

If sinB > 1, then there is no angle B that meets the given information, so no triangle can be formed with the given sides and angle.

If sinB = 1, then we have a right triangle with right angle at B and we can proceed as above for right triangles.

If sinB < 1, then there are two possible measures for angle B, one accute (the inverse sine of the value of sinB) and one obtuse (the supplement of the accute one). The accute angle will give a triangle and we can find the missing information as above now that we have two angles. The obtuse angle may or may not give a triangle--when we attempt to compute the third angle, it may be that A + B is more than a straight angle, leaving no room for angle C. If there is room for an angle C, then there is a second possible triangle determined by the given information and we can find the last missing length as above.

Next Page: Vectors in the Plane | Previous Page: Law of Cosines

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[edit] Vectors in the Plane

In practice, one of the most useful applications of trigonometry is in calculations related to vectors, which are frequently used in Physics. A vector is a quantity which has both magnitude (such as three or eight) and direction (such as north or 30 degrees south of east). It is represented in diagrams by an arrow, often pointing from the origin to a specific point.

A plane vector \vec{A} can be expressed in two ways -- as the sum of a horizontal vector of magnitude Ax and a vertical vector of magnitude Ay, or in terms of its angle θ and magnitude \left|\vec{A}\right| (or simply A). These two methods are called "rectangular" and "polar" respectively.

[edit] Rectangular to Polar conversion

For simplicity, assume \vec{A} is in the first quadrant and has x-component Ax and y-component Ay. Given these components, we want to find the angle θ and the magnitude A.

If we draw all three of these vectors, they form a right triangle. It is easy to see that \tan \theta = \frac{A_y}{A_x}, or \theta = \arctan \frac{A_y}{A_x} (A vector with an angle of zero is defined to be pointing directly to the right.) Furthermore, by the Pythagorean Theorem, A_x\,^2 + A_y\,^2=A^2, or A=\sqrt{A_x\,^2 + A_y\,^2}.

[edit] Polar to Rectangular conversion

This is essentially the same problem as above, but in reverse. Here, θ and A are known and we want to calculate the values of Ax and Ay.

Using the same triangle as above, we can see that \cos \theta = \frac{A_x}{A}, or Ax = Acosθ. Also, \sin \theta = \frac{A_y}{A}, or Ay = Asinθ.

[edit] Review of conversions

  • \theta = \arctan \frac{A_y}{A_x}
  • \left|\vec{A}\right|=A=\sqrt{A_x\,^2 + A_y\,^2}
  • Ax = Acosθ
  • Ay = Asinθ

Next Page: Vectors and Dot Products | Previous Page: Solving Triangles

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[edit] Vectors and Dot Products

Consider the vectors U and V (with respective magnitudes |U| and |V|). If those vectors enclose an angle θ then the dot product of those vectors can be written as:


\mathbf{U}\cdot\mathbf{V} = |\mathbf{U}||\mathbf{V}| \cos(\theta)

If the vectors can be written as:

\mathbf{U} = (U_x, U_y, U_z)
\mathbf{V} = (V_x, V_y, V_z)

then the dot product is given by:

\mathbf{U}\cdot\mathbf{V} = U_x V_x + U_y V_y + U_z V_z

For example,

(1, 2, 3) \cdot (2, 2, 2) = 1 (2) + 2 (2) + 3 (2) = 12.

and

(0, 5, 0) \cdot (4, 0, 0) = 0.

We can interpret the last case by noting that the product is zero because the angle between the two vectors is 90 degrees.

[edit] Trigonometric Form of the Complex Number

z=a+bi= r \left ( \cos \phi \ + i\sin \phi \right )

where

  • i is the Imaginary Number \left (i\ = \sqrt{-1}\right )
  • the modulus r=\operatorname{mod}(z)=|z|=\sqrt{a^2+b^2}
  • the argument \phi=\arg(z) is the angle formed by the complex number on a polar graph with one real axis and one imaginary axis. This can be found using the right angle trigonometry for the trigonometric functions.

This is sometimes abbreviated as r\left ( \cos \phi \ + i\sin \phi \right )=r\operatorname{cis}\phi and it is also the case that r\operatorname{cis}\phi=re^{i\phi} (provided that φ is in radians).


[edit] Trigonometry References

[edit] Trigonometric Unit Circle and Graph Reference

The Unit Circle Unit circle angles.svg

The unit circle is a commonly used tool in trigonometry because it helps the user to remember the special angles and their trigonometric functions. The unit circle is a circle drawn with its center at the origin of a graph(0,0), and with a radius of 1. All angles are measured starting from the x-axis in quadrant one and may go around the unit circle any number of degrees. points on the outside of the circle that are in line with the terminal(ending) sides of the angles are very useful to know, as they give the trigonometric function of the angle through their coordinants. The format is (cos, sin).

[edit] Trigonometric Formula Reference

The principal identities in trigonometry are:

sin2θ + cos2θ = 1

\tan{\theta} = \frac{\sin{\theta}}{\cos{\theta}}

Four trigonometric functions are periodic:

sinθ = sin(θ + 2π)

cosθ = cos(θ + 2π)

cscθ = csc(θ + 2π)

secθ = sec(θ + 2π)

Two trigonometric functions are π periodic:

tanθ = tan(θ + π)

cotθ = cot(θ + π)

Formulas involving sums of angles are as follows:

sin(α + β) = sinαcosβ + cosαsinβ

cos(α + β) = cosαcosβ − sinαsinβ

Substituting β = α gives the double angle formulae

sin(2α) = 2sin(α)cos(α)

cos(2α) = cos2α − sin2α

Substituting sin2α + cos2α = 1 gives

cos(2α) = 2cos2α − 1

cos(2α) = 1 − 2sin2α

sin(3θ) = 3sinθ − 4sin3θ

cos(3θ) = 4cox3θ − 3cosθ

2sin(A)cos(B) = sin(A + B) + sin(AB)


[edit] Trigonometric Identities Reference

[edit] Pythagoras

  1. sin2(x) + cos2(x) = 1
  2. 1 + tan2(x) = sec2(x)
  3. 1 + cot2(x) = csc2(x)

These are all direct consequences of Pythagoras's theorem.

[edit] Sum/Difference of angles

  1. \cos(x\pm y)=\cos(x)\cos(y) \mp \sin(x)\sin(y)
  2. \sin(x\pm y)=\sin(x)\cos(y) \pm \sin(y)\cos(x)
  3. \tan(x\pm y)=\frac{\tan(x) \pm \tan(y)}{1 \mp \tan(x) \tan(y)}

[edit] Product to Sum

  1. 2sin(x)sin(y) = cos(x - y) - cos(x + y)
  2. 2cos(x)cos(y) = cos(x - y) + cos(x + y)
  3. 2sin(x)cos(y) = sin(x - y) + sin(x + y)

[edit] Sum and difference to product

  1. Asin(x) + Bcos(x) = Csin(x + y), where C=\sqrt{A^2+B^2} and y=\pm\arctan(B/A)
  2. \sin\alpha+\sin\beta=2\sin\frac{\alpha+\beta}{2}\cos\frac{\alpha-\beta}{2}
  3. \sin\alpha-\sin\beta=2\cos\frac{\alpha+\beta}{2}\sin\frac{\alpha-\beta}{2}
  4. \cos\alpha+\cos\beta=2\cos\frac{\alpha+\beta}{2}\cos\frac{\alpha-\beta}{2}
  5. \cos\alpha-\cos\beta=-2\sin\frac{\alpha+\beta}{2}\sin\frac{\alpha-\beta}{2}

[edit] Double angle

  1. cos(2x) = cos2(x) − sin2(x) = 2cos2(x) − 1 = 1 − 2sin2(x)
  2. sin(2x) = 2sin(x)cos(x)
  3. \tan(2x)=\frac{2\tan(x)}{1- \tan^2(x)}

These are all direct consequences of the sum/difference formulae

[edit] Half angle

  1. \cos(\frac{x}{2})=\pm\sqrt{\frac{1+\cos(x)}{2}}
  2. \sin(\frac{x}{2})=\pm\sqrt{\frac{1-\cos(x)}{2}}
  3. \tan(\frac{x}{2})=\frac{1-\cos(x)}{\sin(x)}=\frac{\sin(x)}{1+\cos(x)}=\pm\sqrt{\frac{1-\cos(x)}{1+\cos(x)}}

In cases with \pm, the sign of the result must be determined from the value of \frac{x}{2}. These derive from the cos(2x) formulae.

[edit] Power Reduction

  1. \sin^2\theta=\frac{1-\cos2\theta}{2}
  2. \cos^2\theta=\frac{1+\cos2\theta}{2}
  3. \tan^2\theta=\frac{1-\cos2\theta}{1+\cos2\theta}

[edit] Even/Odd

  1. sin( − θ) = − sin(θ)
  2. cos( − θ) = cos(θ)
  3. tan( − θ) = − tan(θ)
  4. csc( − θ) = − csc(θ)
  5. sec( − θ) = sec(θ)
  6. cot( − θ) = − cot(θ)

[edit] Calculus

  1. \frac{d}{dx}[\sin x] = \cos x
  2. \frac{d}{dx}[\cos x] = -\sin x
  3. \frac{d}{dx}[\tan x] = \sec^{2} x
  4. \frac{d}{dx}[\sec x] = \sec x \tan x
  5. \frac{d}{dx}[\csc x] = -\csc x \cot x
  6. \frac{d}{dx}[\cot x] = -\csc^{2} x

[edit] Natural Trigonometric Functions of Primary Angles

Note: Some values in the table are given in forms that include a radical in the denominator — this is done both to simplify recognition of reciprocal pairs and because the form given in the table is simpler in some sense. Note also that all absolute values of trigonometric functions for remarquable points that are listed in this table are contained in the first quadrant (from 0 to 90° or π / 2 radians, inclusive); all others are deduced by simple symetries with the horizontal or vertical axis, or by swapping axis (on the trigonometric circle) so that one trigonometric function is also swapped with its co-function.

\theta\, (positive) \sin\theta\, \cos\theta\, \tan\theta\, \cot\theta\, \sec\theta\, \csc\theta\, \theta\, (negative)
(degrees) (radians) (degrees) (radians)
0\, 0\, 1\, 0\, not
defined
1\, not
defined
−360° -2\pi\,
15° \frac{\pi}{12}\, \frac{\sqrt{6}-\sqrt{2}}{4}\, \frac{\sqrt{6}+\sqrt{2}}{4}\, 2-\sqrt{3}\, 2+\sqrt{3}\, \frac{4}{\sqrt{6}+\sqrt{2}}\, \frac{4}{\sqrt{6}-\sqrt{2}}\, −345° -\frac{13\pi}{12}\,
22.5° \frac{\pi}{8}\, \frac{\sqrt{2-\sqrt{2}}}{2}\, \frac{\sqrt{2+\sqrt{2}}}{2}\, \sqrt{2}-1\, \sqrt{2}+1\, \frac{2}{\sqrt{2+\sqrt{2}}}\, \frac{2}{\sqrt{2-\sqrt{2}}}\, −337.5° -\frac{15\pi}{8}\,
30° \frac{\pi}{6}\, \frac{1}{2}\, \frac{\sqrt{3}}{2}\, \frac{1}{\sqrt{3}}\, \sqrt{3}\, \frac{2}{\sqrt{3}}\, 2\, −330° -\frac{11\pi}{6}\,
45° \frac{\pi}{4}\, \frac{1}{\sqrt{2}}\, \frac{1}{\sqrt{2}}\, 1\, 1\, \sqrt{2}\, \sqrt{2}\, −315° -\frac{7\pi}{4}\,
60° \frac{\pi}{3}\, \frac{\sqrt{3}}{2}\, \frac{1}{2}\, \sqrt{3}\, \frac{1}{\sqrt{3}}\, 2\, \frac{2}{\sqrt{3}}\, −300° -\frac{5\pi}{3}\,
67.5° \frac{3\pi}{8}\, \frac{\sqrt{2+\sqrt{2}}}{2}\, \frac{\sqrt{2-\sqrt{2}}}{2}\, \sqrt{2}+1\, \sqrt{2}-1\, \frac{2}{\sqrt{2-\sqrt{2}}}\, \frac{2}{\sqrt{2+\sqrt{2}}}\, −292.5° -\frac{11\pi}{8}\,
75° \frac{5\pi}{12}\, \frac{\sqrt{6}+\sqrt{2}}{4}\, \frac{\sqrt{6}-\sqrt{2}}{4}\, 2+\sqrt{3}\, 2-\sqrt{3}\, \frac{4}{\sqrt{6}-\sqrt{2}}\, \frac{4}{\sqrt{6}+\sqrt{2}}\, −285° -\frac{19\pi}{12}\,
90° \frac{\pi}{2}\, 1\, 0\, not
defined
0\, not
defined
1\, −270° -\frac{3\pi}{2}\,
105° \frac{7\pi}{12}\, \frac{\sqrt{6}+\sqrt{2}}{4}\, -\frac{\sqrt{6}-\sqrt{2}}{4}\, -2-\sqrt{3}\, -2+\sqrt{3}\, -\frac{4}{\sqrt{6}-\sqrt{2}}\, \frac{4}{\sqrt{6}+\sqrt{2}}\, −255° -\frac{17\pi}{12}\,
112.5° \frac{5\pi}{8}\, \frac{\sqrt{2+\sqrt{2}}}{2}\, -\frac{\sqrt{2-\sqrt{2}}}{2}\, -\sqrt{2}-1\, -\sqrt{2}+1\, -\frac{2}{\sqrt{2-\sqrt{2}}}\, \frac{2}{\sqrt{2+\sqrt{2}}}\, −247.5° -\frac{11\pi}{8}\,
120° \frac{2\pi}{3}\, \frac{\sqrt{3}}{2}\, -\frac{1}{2}\, -\sqrt{3}\, -\frac{1}{\sqrt{3}}\, -2\, \frac{2}{\sqrt{3}}\, −240° -\frac{4\pi}{3}\,
135° \frac{3\pi}{4}\, \frac{1}{\sqrt{2}}\, -\frac{1}{\sqrt{2}}\, -1\, -1\, -\sqrt{2}\, \sqrt{2}\, −225° -\frac{5\pi}{4}\,
150° \frac{5\pi}{6}\, \frac{1}{2}\, -\frac{\sqrt{3}}{2}\, -\frac{1}{\sqrt{3}}\, -\sqrt{3}\, -\frac{2}{\sqrt{3}}\, 2\, −210° -\frac{7\pi}{6}\,
157.5° \frac{7\pi}{8}\, \frac{\sqrt{2-\sqrt{2}}}{2}\, -\frac{\sqrt{2+\sqrt{2}}}{2}\, -\sqrt{2}+1\, -\sqrt{2}-1\, -\frac{2}{\sqrt{2+\sqrt{2}}}\, \frac{2}{\sqrt{2-\sqrt{2}}}\, −202.5° -\frac{9\pi}{8}\,
165° \frac{11\pi}{12}\, \frac{\sqrt{6}-\sqrt{2}}{4}\, -\frac{\sqrt{6}+\sqrt{2}}{4}\, -2+\sqrt{3}\, -2-\sqrt{3}\, -\frac{4}{\sqrt{6}+\sqrt{2}}\, \frac{4}{\sqrt{6}-\sqrt{2}}\, −195° -\frac{13\pi}{12}\,
180° \pi\, 0\, -1\, 0\, not
defined
-1\, not
defined
−180° -\pi\,
195° \frac{13\pi}{12}\, -\frac{\sqrt{6}-\sqrt{2}}{4}\, -\frac{\sqrt{6}+\sqrt{2}}{4}\, 2-\sqrt{3}\, 2+\sqrt{3}\, -\frac{4}{\sqrt{6}+\sqrt{2}}\, -\frac{4}{\sqrt{6}-\sqrt{2}}\, −165° -\frac{11\pi}{12}\,
202.5° \frac{9\pi}{8}\, -\frac{\sqrt{2-\sqrt{2}}}{2}\, -\frac{\sqrt{2+\sqrt{2}}}{2}\, \sqrt{2}-1\, \sqrt{2}+1\, -\frac{2}{\sqrt{2+\sqrt{2}}}\, -\frac{2}{\sqrt{2-\sqrt{2}}}\, −157.5° -\frac{7\pi}{8}\,
210° \frac{7\pi}{6}\, -\frac{1}{2}\, -\frac{\sqrt{3}}{2}\, \frac{1}{\sqrt{3}}\, \sqrt{3}\, -\frac{2}{\sqrt{3}}\, -2\, −150° -\frac{5\pi}{6}\,
225° \frac{5\pi}{4}\, -\frac{1}{\sqrt{2}}\, -\frac{1}{\sqrt{2}}\, 1\, 1\, -\sqrt{2}\, -\sqrt{2}\, −135° -\frac{3\pi}{4}\,
240° \frac{4\pi}{3}\, -\frac{\sqrt{3}}{2}\, -\frac{1}{2}\, \sqrt{3}\, \frac{1}{\sqrt{3}}\, -2\, -\frac{2}{\sqrt{3}}\, −120° -\frac{2\pi}{3}\,
247.5° \frac{11\pi}{8}\, -\frac{\sqrt{2+\sqrt{2}}}{2}\, -\frac{\sqrt{2-\sqrt{2}}}{2}\, \sqrt{2}+1\, \sqrt{2}-1\, -\frac{2}{\sqrt{2-\sqrt{2}}}\, -\frac{2}{\sqrt{2+\sqrt{2}}}\, −112.5° -\frac{5\pi}{8}\,
255° \frac{17\pi}{12}\, -\frac{\sqrt{6}+\sqrt{2}}{4}\, -\frac{\sqrt{6}-\sqrt{2}}{4}\, 2+\sqrt{3}\, 2-\sqrt{3}\, -\frac{4}{\sqrt{6}-\sqrt{2}}\, -\frac{4}{\sqrt{6}+\sqrt{2}}\, −105° -\frac{7\pi}{12}\,
270° \frac{3\pi}{2}\, -1\, 0\, not
defined
0\, not
defined
-1\, −90° -\frac{\pi}{2}\,
285° \frac{19\pi}{12}\, -\frac{\sqrt{6}+\sqrt{2}}{4}\, \frac{\sqrt{6}-\sqrt{2}}{4}\, -2-\sqrt{3}\, -2+\sqrt{3}\, \frac{4}{\sqrt{6}-\sqrt{2}}\, -\frac{4}{\sqrt{6}+\sqrt{2}}\, −75° -\frac{5\pi}{12}\,
292.5° \frac{11\pi}{8}\, -\frac{\sqrt{2+\sqrt{2}}}{2}\, \frac{\sqrt{2-\sqrt{2}}}{2}\, -\sqrt{2}-1\, -\sqrt{2}+1\, \frac{2}{\sqrt{2-\sqrt{2}}}\, -\frac{2}{\sqrt{2+\sqrt{2}}}\, −67.5° -\frac{3\pi}{8}\,
300° \frac{5\pi}{3}\, -\frac{\sqrt{3}}{2}\, \frac{1}{2}\, -\sqrt{3}\, -\frac{1}{\sqrt{3}}\, 2\, -\frac{2}{\sqrt{3}}\, −60° -\frac{\pi}{3}\,
315° \frac{7\pi}{4}\, -\frac{1}{\sqrt{2}}\, \frac{1}{\sqrt{2}}\, -1\, -1\, \sqrt{2}\, -\sqrt{2}\, −45° -\frac{\pi}{4}\,
330° \frac{11\pi}{6}\, -\frac{1}{2}\, \frac{\sqrt{3}}{2}\, -\frac{1}{\sqrt{3}}\, -\sqrt{3}\, \frac{2}{\sqrt{3}}\, -2\, −30° -\frac{\pi}{6}\,
337.5° \frac{15\pi}{8}\, -\frac{\sqrt{2-\sqrt{2}}}{2}\, \frac{\sqrt{2+\sqrt{2}}}{2}\, -\sqrt{2}+1\, -\sqrt{2}-1\, \frac{2}{\sqrt{2+\sqrt{2}}}\, -\frac{2}{\sqrt{2-\sqrt{2}}}\, −22.5° -\frac{\pi}{8}\,
345° \frac{13\pi}{12}\, -\frac{\sqrt{6}-\sqrt{2}}{4}\, \frac{\sqrt{6}+\sqrt{2}}{4}\, -2+\sqrt{3}\, -2-\sqrt{3}\, \frac{4}{\sqrt{6}+\sqrt{2}}\, -\frac{4}{\sqrt{6}-\sqrt{2}}\, −15° -\frac{\pi}{12}\,
360° 2\pi\, 0\, 1\, 0\, not
defined
1\, not
defined
0\,

Notice that for certain values of x\,, the tangent, cotangent, secant, and cosecant functions are undefined. This is because these functions are defined as \frac{\sin x}{\cos x}\,, \frac{\cos x}{\sin x}\,, \frac1{\cos x}\,, and \frac1{\sin x}\,, respectively. Since an expression is undefined if it is divided by zero, the functions are therefore undefined at angle measures where the denominator (the sine or cosine of x\,, depending on the trigonometric function) is equal to zero. Take, for example, the tangent function. If the tangent function is analyzed at 90 degrees (\pi/2\, radians), the function is then equivalent to \frac{\sin(\pi/2)}{\cos(\pi/2)}\,, or \frac10\,, which is an undefined value.

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[edit] License

[edit] GNU Free Documentation License

Version 1.2, November 2002

Copyright (C) 2000,2001,2002  Free Software Foundation, Inc.
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