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 $$\varphi_2$$ 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:

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

which is identical to the "Cosine Double Angle Sum Formula": $$2 cos(\delta)^2 - 1 = cos(2\delta)\,$$

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:

$$cos(\theta_1+\theta_2) = cos(\theta_1)cos(\theta_2) - sin(\theta_1)sin(\theta_2)\,$$ $$\Rightarrow cos(\theta_1+\theta_2)^2 = (cos(\theta_1)cos(\theta_2) - sin(\theta_1)sin(\theta_2))^2\,$$ $$\Rightarrow 1 - cos(\theta_1+\theta_2)^2 = 1 - (cos(\theta_1)cos(\theta_2) - sin(\theta_1)sin(\theta_2))^2\,$$ $$\Rightarrow sin(\theta_1+\theta_2)^2    = 1 - (cos(\theta_1)^2cos(\theta_2)^2 + sin(\theta_1)^2sin(\theta_2)^2 - 2cos(\theta_1)cos(\theta_2)sin(\theta_1)sin(\theta_2))\,$$ -- 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

$$= 1 - (cos(\theta_1)^2-cos(\theta_1)^2sin(\theta_2)^2 + sin(\theta_1)^2-sin(\theta_1)^2cos(\theta_2)^2 - 2cos(\theta_1)cos(\theta_2)sin(\theta_1)sin(\theta_2))\,$$ -- Multiplied out

$$= 1 - (1 -cos(\theta_1)^2sin(\theta_2)^2 -sin(\theta_1)^2cos(\theta_2)^2 - 2cos(\theta_1)cos(\theta_2)sin(\theta_1)sin(\theta_2))\,$$ -- Carefully selected Pythagoras

$$= cos(\theta_1)^2sin(\theta_2)^2+sin(\theta_1)^2cos(\theta_2)^2 + 2cos(\theta_1)cos(\theta_2)sin(\theta_1)sin(\theta_2)\,$$ -- 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: $$sin \left ( \frac{\theta}{2} \right ) = \sqrt{\frac{1}{2}}\sqrt{1 - \sqrt{1 - sin(\theta)^2}}$$