Sine and Cosine of Complementary Angles

Theorem
Let $\alpha$ and $\beta$ be complementary angles.

Then:
 * $\sin \alpha = \cos \beta$

where $\sin$ and $\cos$ are sine and cosine.

Proof 1
Similarly and alternatively:

Proof 2

 * From Sine and Cosine are Periodic on Reals, we have:
 * $\sin \left({x + \dfrac \pi 2}\right) = \cos x$
 * $\cos \left({x + \dfrac \pi 2}\right) = -\sin x$


 * Also from Sine and Cosine are Periodic on Reals, we have:
 * $\sin \left({x + \pi}\right) = \cos \left({x + \dfrac \pi 2}\right) = -\sin x$
 * $\cos \left({x + \pi}\right) = -\sin \left({x + \dfrac \pi 2}\right) = -\cos x$


 * From Sine Function is Odd, we have:
 * $\sin \left({x + \dfrac \pi 2}\right) = - \sin \left({- x - \dfrac \pi 2}\right)$


 * From Cosine Function is Even, we have:
 * $\cos \left({x + \dfrac \pi 2}\right) = \cos \left({- x - \dfrac \pi 2}\right)$

So:

... and: