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?

The answer is yes.

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 ${\theta }_1\ne {\theta }_2$, 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 ${\phi }_2$ 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) ${\theta }_4+(\pi -\frac{\pi }{2}-{\theta }_1-{\theta }_2)={\phi }_2=\pi -\frac{\pi }{2}-{\theta }_2\Rightarrow {\theta }_4={\theta }_1$. 

The length of side $b_4$ is $cos({\phi }_2)cos({\phi }_4)=cos({\phi }_2)cos({\phi }_1)$ as ${\theta }_4={\theta }_1$. Thus point $g$ is located at length:

$cos({\theta }_1+{\theta }_2)=cos({\theta }_1)cos({\theta }_2)-cos({\phi }_1)cos({\phi }_2),$ where ${\theta }_1+{\phi }_1={\theta }_2+{\phi }_2=\frac{\pi }{2}$

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:

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

From the theorem of Pythagoras we know that:

  $a^2+b^2=c^2$

in this case:

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

Substituting into (I) gives:

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

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:

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

to:

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

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

 $cos({\theta }_1+{\theta }_2)=cos({\theta }_1)cos({\theta }_2)-cos({\phi }_1)cos({\phi }_2)$ where ${\theta }_1+{\phi }_1={\theta }_2+{\phi }_2=\frac{\pi }{2}$

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:

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