Cosine of Complement equals Sine

Theorem

 * $\sin \theta = \cos \left({\dfrac \pi 2 - \theta}\right)$

where $\dfrac \pi 2 - \theta$ is the complement of $\theta$.

Proof

 * From Sine and Cosine are Periodic on Reals, we have $\cos \left({x + \dfrac \pi 2}\right) = -\sin x$;
 * Also from Sine and Cosine are Periodic on Reals, we have $\cos \left({x + \pi}\right) = -\sin \left({x + \dfrac \pi 2}\right) = -\cos x$;
 * From Basic Properties of Cosine Function, we have $\cos \left({x + \dfrac \pi 2}\right) = \cos \left({- x - \dfrac \pi 2}\right)$.

So:

Note
Compare Cosine equals Sine of Complement.