All of the trigonometric functions of an angle θ can be constructed geometrically in terms of a unit circle centered at O. Given that:
Sine function
![f(\theta) = sin(\theta)](https://tex.z-dn.net/?f=f%28%5Ctheta%29%20%3D%20sin%28%5Ctheta%29)
being
Cosine function
![f(\theta) = cos(\theta)](https://tex.z-dn.net/?f=f%28%5Ctheta%29%20%3D%20cos%28%5Ctheta%29)
being
We will demonstrate the identity above. First of all, we need to square each equation, so:
![sin^{2}(\theta)= v^{2}](https://tex.z-dn.net/?f=sin%5E%7B2%7D%28%5Ctheta%29%3D%20v%5E%7B2%7D)
![cos^{2}(\theta)=u^{2}](https://tex.z-dn.net/?f=cos%5E%7B2%7D%28%5Ctheta%29%3Du%5E%7B2%7D)
Adding these two equations:
![sin^{2}(\theta)+cos^{2}(\theta)=v ^{2}+u^{2}](https://tex.z-dn.net/?f=sin%5E%7B2%7D%28%5Ctheta%29%2Bcos%5E%7B2%7D%28%5Ctheta%29%3Dv%20%5E%7B2%7D%2Bu%5E%7B2%7D)
But as shown in the figure, using Pythagorean theorem
![v^{2}+u^{2}](https://tex.z-dn.net/?f=v%5E%7B2%7D%2Bu%5E%7B2%7D)
is always equal to 1, then:
![sin^{2}(\theta)+cos^{2}(\theta)=1](https://tex.z-dn.net/?f=sin%5E%7B2%7D%28%5Ctheta%29%2Bcos%5E%7B2%7D%28%5Ctheta%29%3D1)
The relation to right triangles is that:
<span>The hypotenuse is always equal to 1
</span>The opposite side is equal to
![sin(\theta)](https://tex.z-dn.net/?f=sin%28%5Ctheta%29)
The adjacent side is equal to