Sine of Complement equals Cosine

Theorem

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

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

Proof

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

So:

Note
Compare Sine equals Cosine of Complement.