Sum of Arcsine and Arccosine

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

Then:
 * $\arcsin x + \arccos x = \dfrac \pi 2$

where $\arcsin$ and $\arccos$ denote arcsine and arccosine respectively.

Proof
Let $y \in \R$ such that:
 * $\exists x \in \closedint {-1} 1: x = \map \cos {y + \dfrac \pi 2}$

Then:

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

Then we can write:
 * $-y = \arcsin x$

But then:
 * $\map \cos {y + \dfrac \pi 2} = x$

Now since $-\dfrac \pi 2 \le y \le \dfrac \pi 2$ it follows that:
 * $0 \le y + \dfrac \pi 2 \le \pi$

Hence:
 * $y + \dfrac \pi 2 = \arccos x$

That is:
 * $\dfrac \pi 2 = \arccos x + \arcsin x$

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


 * $\map {D_x} {\arcsin x + \arccos x} = \dfrac 1 {\sqrt {1 - x^2} } + \dfrac {-1} {\sqrt {1 - x^2}} = 0$

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