# Trigonometry/For Enthusiasts/Doing without Sine

## The Idea

We know that:

${\displaystyle \sin ^{2}\theta =1-\cos ^{2}\theta \,}$

So do we really need the ${\displaystyle \sin \,}$ function?

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

We don't need to have one geometric argument for ${\displaystyle \cos(a+b)\,}$ and then do another geometric argument for ${\displaystyle \sin(a+b)\,}$. We could get our formulas for ${\displaystyle \sin \,}$ directly from formulas for ${\displaystyle \cos \,}$

## Angle Addition and Subtraction formulas

To find a formula for ${\displaystyle \cos(\theta _{1}+\theta _{2})\,}$ in terms of ${\displaystyle \cos \left(\theta _{1}\right)}$ and ${\displaystyle \cos \left(\theta _{2}\right)}$: construct two different right angle triangles each drawn with side ${\displaystyle c\,}$ having the same length of one, but with ${\displaystyle \theta _{1}\neq \theta _{2}}$, and therefore angle ${\displaystyle \psi _{1}\neq \psi _{2}}$. Scale up triangle two so that side ${\displaystyle a_{2}\,}$ is the same length as side ${\displaystyle c_{1}\,}$. Place the triangles so that side ${\displaystyle c_{1}\,}$ is coincidental with side ${\displaystyle a_{2}\,}$, and the angles ${\displaystyle \theta _{1}\,}$ and ${\displaystyle \theta _{2}\,}$ are juxtaposed to form angle ${\displaystyle \theta _{3}=\theta _{1}+\theta _{2}\,}$ at the origin. The circumference of the circle within which triangle two is embedded (circle 2) crosses side ${\displaystyle a_{1}\,}$ at point ${\displaystyle g\,}$, allowing a third right angle to be drawn from angle ${\displaystyle \varphi _{2}}$ to point ${\displaystyle g\,}$ . Now reset the scale of the entire figure so that side ${\displaystyle c_{2}\,}$ is considered to be of length 1. Side ${\displaystyle a_{2}\,}$ coincidental with side ${\displaystyle c_{1}\,}$ will then be of length ${\displaystyle \cos \left(\theta _{1}\right)}$, and so side ${\displaystyle a_{1}\,}$ will be of length ${\displaystyle \cos \left(\theta _{1}\right)\cdot \cos \left(\theta _{2}\right)}$ in which length lies point ${\displaystyle g\,}$. Draw a line parallel to line ${\displaystyle a_{1}\,}$ 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) ${\displaystyle \theta _{4}+(\pi -{\frac {\pi }{2}}-\theta _{1}-\theta _{2})=\varphi _{2}=\pi -{\frac {\pi }{2}}-\theta _{2}\Rightarrow \theta _{4}=\theta _{1}}$.


The length of side ${\displaystyle b_{4}\,}$ is ${\displaystyle cos(\varphi _{2})cos(\varphi _{4})=cos(\varphi _{2})cos(\varphi _{1})}$ as ${\displaystyle \theta _{4}\,=\theta _{1}\,}$. Thus point ${\displaystyle g\,}$ is located at length:

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

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:

   ${\displaystyle cos(2\theta )=cos(\theta +\theta )=cos(\theta )cos(\theta )-cos(\varphi )cos(\varphi )=cos(\theta )^{2}-cos(\varphi )^{2}}$           (I)
where ${\displaystyle \theta +\varphi ={\frac {\pi }{2}}}$


From the theorem of Pythagoras we know that:

  ${\displaystyle a^{2}+b^{2}=c^{2}\,}$


in this case:

   ${\displaystyle cos(\theta )^{2}+cos(\varphi )^{2}=1^{2}}$
${\displaystyle \Rightarrow cos(\varphi )^{2}=1-cos(\theta )^{2}}$
where ${\displaystyle \theta +\varphi ={\frac {\pi }{2}}}$


Substituting into (I) gives:

   ${\displaystyle cos(2\theta )=cos(\theta )^{2}-cos(\varphi )^{2}}$
${\displaystyle =cos(\theta )^{2}-(1-cos(\theta )^{2})\,}$
${\displaystyle =2cos(\theta )^{2}-1\,}$
where ${\displaystyle \theta +\varphi ={\frac {\pi }{4}}}$


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

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


## Pythagorean identity

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

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


to:

    ${\displaystyle cos(\theta )^{2}+sin(\theta )^{2}=1\,}$


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

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


to:

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


## Sine Formulas

The price we have to pay for the notational convenience of this new function ${\displaystyle sin()\,}$ 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 ${\displaystyle cos()\,}$ form and selectively replacing ${\displaystyle cos(\theta )^{2}\,}$ by ${\displaystyle 1-sin(\theta )^{2}\,}$ and then using algebra to simplify the resulting equation. Applying this technique to the "Cosine Angle Sum Formula" produces:

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

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

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

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

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

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


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

${\displaystyle \Rightarrow sin(\theta _{1}+\theta _{2})=cos(\theta _{1})sin(\theta _{2})+sin(\theta _{1})cos(\theta _{2})}$


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

${\displaystyle cos\left({\frac {\theta }{2}}\right)={\sqrt {\frac {1+cos(\theta )}{2}}}}$

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

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


So far so good, but we still have a ${\displaystyle {\frac {cos(\theta )}{2}}}$ to get rid of. Use Pythagoras again to get the "Sine Half Angle Formula":

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


or perhaps a little more legibly as:

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