Sum of Arcsine and Arccosine

Theorem
Let $$x \in \R$$ be a real number such that $$-1 \le x \le 1$$.

Let:
 * $$\arcsin x$$ be the arcsine of $$x$$.
 * $$\arccos x$$ be the arccosine of $$x$$.

Then $$\arcsin x + \arccos x = \frac \pi 2$$.

Proof
Let $$y \in \R$$.

Then $$\cos \left({y + \frac \pi 2}\right) = - \sin y = \sin \left({-x}\right)$$.

Suppose $$-\frac \pi 2 \le y \le \frac \pi 2$$.

Then we can write $$- x = \arcsin y$$.

But then $$\cos \left({y + \frac \pi 2}\right) = x$$.

Now since $$-\frac \pi 2 \le y \le \frac \pi 2$$ it follows that $$0 \le y + \frac \pi 2 \le \pi$$.

Hence $$y + \frac \pi 2 = \arccos x$$.

That is, $$\frac \pi 2 = \arccos x + \arcsin x$$.

Note
Note that from Derivative of Arcsine Function and Derivative of Arccosine Function, we have:


 * $$D_x \left({\arcsin x + \arccos x}\right) = \frac 1 {\sqrt {1 - x^2}} + \frac {-1} {\sqrt {1 - x^2}} = 0$$

which is what (from Derivative of Constant) we would expect.