# Trigonometry/For Enthusiasts/Doing without Sine

## The Idea

We know that:

$\sin ^{2}\theta =1-\cos ^{2}\theta \,$ So do we really need the $\sin \,$ function?

Or put another way, could we have worked out all our interesting formulas for things like $\cos(a+b)\,$ in terms just of $\cos \,$ and then derived every formula that has a $\sin \,$ in it from that?

We don't need to have one geometric argument for $\cos(a+b)\,$ and then do another geometric argument for $\sin(a+b)\,$ . We could get our formulas for $\sin \,$ directly from formulas for $\cos \,$ ## 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 $\theta _{1}\neq \theta _{2}$ , and therefore angle $\psi _{1}\neq \psi _{2}$ . 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 $b_{4}\,$ is $cos(\varphi _{2})cos(\varphi _{4})=cos(\varphi _{2})cos(\varphi _{1})$ as $\theta _{4}\,=\theta _{1}\,$ . Thus point $g\,$ is located at length:

$cos(\theta _{1}+\theta _{2})=cos(\theta _{1})cos(\theta _{2})-cos(\varphi _{1})cos(\varphi _{2}),$ where $\theta _{1}+\varphi _{1}=\theta _{2}+\varphi _{2}={\frac {\pi }{2}}$ giving us the "Cosine Angle Sum Formula".

### Proof that angle sum formula and double angle formula are consistent

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

   $cos(2\theta )=cos(\theta +\theta )=cos(\theta )cos(\theta )-cos(\varphi )cos(\varphi )=cos(\theta )^{2}-cos(\varphi )^{2}$ (I)
where $\theta +\varphi ={\frac {\pi }{2}}$ From the theorem of Pythagoras we know that:

  $a^{2}+b^{2}=c^{2}\,$ in this case:

   $cos(\theta )^{2}+cos(\varphi )^{2}=1^{2}$ $\Rightarrow cos(\varphi )^{2}=1-cos(\theta )^{2}$ where $\theta +\varphi ={\frac {\pi }{2}}$ Substituting into (I) gives:

   $cos(2\theta )=cos(\theta )^{2}-cos(\varphi )^{2}$ $=cos(\theta )^{2}-(1-cos(\theta )^{2})\,$ $=2cos(\theta )^{2}-1\,$ where $\theta +\varphi ={\frac {\pi }{4}}$ which is identical to the "Cosine Double Angle Sum Formula":

   $2cos(\delta )^{2}-1=cos(2\delta )\,$ ## 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:

    $cos(\theta )^{2}+cos(\varphi )^{2}=1^{2}$ where $\theta +\varphi ={\frac {\pi }{2}}$ to:

    $cos(\theta )^{2}+sin(\theta )^{2}=1\,$ We can also restate the "Cosine Angle Sum Formula" from:

 $cos(\theta _{1}+\theta _{2})=cos(\theta _{1})cos(\theta _{2})-cos(\varphi _{1})cos(\varphi _{2})$ where $\theta _{1}+\varphi _{1}=\theta _{2}+\varphi _{2}={\frac {\pi }{2}}$ to:

 $cos(\theta _{1}+\theta _{2})=cos(\theta _{1})cos(\theta _{2})-sin(\theta _{1})sin(\theta _{2})\,$ ## 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 $cos(\theta )^{2}\,$ by $1-sin(\theta )^{2}\,$ and then using algebra to simplify the resulting equation. Applying this technique to the "Cosine Angle Sum Formula" produces:

   $cos(\theta _{1}+\theta _{2})=cos(\theta _{1})cos(\theta _{2})-sin(\theta _{1})sin(\theta _{2})\,$ $\Rightarrow cos(\theta _{1}+\theta _{2})^{2}=(cos(\theta _{1})cos(\theta _{2})-sin(\theta _{1})sin(\theta _{2}))^{2}\,$ $\Rightarrow 1-cos(\theta _{1}+\theta _{2})^{2}=1-(cos(\theta _{1})cos(\theta _{2})-sin(\theta _{1})sin(\theta _{2}))^{2}\,$ $\Rightarrow sin(\theta _{1}+\theta _{2})^{2}=1-(cos(\theta _{1})^{2}cos(\theta _{2})^{2}+sin(\theta _{1})^{2}sin(\theta _{2})^{2}-2cos(\theta _{1})cos(\theta _{2})sin(\theta _{1})sin(\theta _{2}))\,$ -- Pythagoras on left, multiply out right hand side

                     $=1-(cos(\theta _{1})^{2}(1-sin(\theta _{2})^{2})+sin(\theta _{1})^{2}(1-cos(\theta _{2})^{2})-2cos(\theta _{1})cos(\theta _{2})sin(\theta _{1})sin(\theta _{2}))\,$ -- Carefully selected Pythagoras again on the left hand side

                     $=1-(cos(\theta _{1})^{2}-cos(\theta _{1})^{2}sin(\theta _{2})^{2}+sin(\theta _{1})^{2}-sin(\theta _{1})^{2}cos(\theta _{2})^{2}-2cos(\theta _{1})cos(\theta _{2})sin(\theta _{1})sin(\theta _{2}))\,$ -- Multiplied out

                     $=1-(1-cos(\theta _{1})^{2}sin(\theta _{2})^{2}-sin(\theta _{1})^{2}cos(\theta _{2})^{2}-2cos(\theta _{1})cos(\theta _{2})sin(\theta _{1})sin(\theta _{2}))\,$ -- Carefully selected Pythagoras

                     $=cos(\theta _{1})^{2}sin(\theta _{2})^{2}+sin(\theta _{1})^{2}cos(\theta _{2})^{2}+2cos(\theta _{1})cos(\theta _{2})sin(\theta _{1})sin(\theta _{2})\,$ -- Algebraic simplification

                     $=(cos(\theta _{1})sin(\theta _{2})+sin(\theta _{1})cos(\theta _{2}))^{2}\,$ taking the square root of both sides produces the "Sine Angle Sum Formula"

$\Rightarrow sin(\theta _{1}+\theta _{2})=cos(\theta _{1})sin(\theta _{2})+sin(\theta _{1})cos(\theta _{2})$ We can use a similar technique to find the "Sine Half Angle Formula" from the "Cosine Half Angle Formula":

$cos\left({\frac {\theta }{2}}\right)={\sqrt {\frac {1+cos(\theta )}{2}}}$ We know that $1-cos\left({\frac {\theta }{2}}\right)^{2}=sin\left({\frac {\theta }{2}}\right)^{2}$ , so squaring both sides of the "Cosine Half Angle Formula" and subtracting from one:

 $\Rightarrow 1-cos\left({\frac {\theta }{2}}\right)^{2}=1-{\frac {1+cos(\theta )}{2}}$ $\Rightarrow sin\left({\frac {\theta }{2}}\right)^{2}={\frac {1-cos(\theta )}{2}}$ $\Rightarrow sin\left({\frac {\theta }{2}}\right)={\sqrt {\frac {1-cos(\theta )}{2}}}$ So far so good, but we still have a ${\frac {cos(\theta )}{2}}$ to get rid of. Use Pythagoras again to get the "Sine Half Angle Formula":

         $sin\left({\frac {\theta }{2}}\right)={\sqrt {\frac {1-{\sqrt {1-sin(\theta )^{2}}}}{2}}}$ or perhaps a little more legibly as:

         $sin\left({\frac {\theta }{2}}\right)={\sqrt {\frac {1}{2}}}{\sqrt {1-{\sqrt {1-sin(\theta )^{2}}}}}$ 