Inverse trig function equation How would you suggest I go about solving this question? I've been thinking about it for ages and nothing comes to mind. 
$$\arcsin x + \arccos x = \frac{\pi}{2}$$ 
 A: METHODOLOGY $1$:  Calculus Based 
Let $f(x)=\arcsin(x)+\arccos(x)$.  Note that $f'(x)=0$.  Therefore, $f(x)$ is constant.
Since $f(0)=\pi/2$, $f(x)=\pi/2$ for all $x$.

METHODOLOGY $2$:  Non-Calculus Based
Alternatively, $\sin(f(x))=x^2+\sqrt{1-x^2}\sqrt{1-x^2}=1$.  Therefore, $f(x)=\pi/2+2n\pi$ for some $n$.  
Noting that $|f(x)|\le \pi$, we conclude $f(x)=\pi/2$.
A: Hint: Define $f(x)=\arcsin(x)+\arccos(x)$. Find $f'(x)$, see that it is $0$ and conclude that $f(x)=\text{const.}=c$. Plug in x=0 to find the value of $c$, which happens to be $\pi/2$.

Another method goes like this: Let $g(x)=\arcsin(x)$ and $h(x)=\pi/2-\arccos(x)$. Now take the sine of both equations:
$\sin(g)=x$ and $\sin(h)=\sin(\pi/2-\arccos(x))$. The second equation can be transformed by using the complementary angle formula for $\sin(\pi/2-x)=\cos(x)$ to result in $\sin(h)=\cos(\arccos(x))=x$. Hence, $\sin(g)=\sin(h)$ or $g=h+2\pi k$. By plugging in a specific value for $x=0$ you will see that $k=0$.
A: $$\arccos x=\frac\pi2-\arcsin x$$
Then, taking the cosine,
$$\cos(\arccos x)=\cos\left(\frac\pi2-\arcsin x\right)=\sin(\arcsin x),$$
and by the definition of these functions, for all $-1\le x\le1$,
$$x=x.$$

Alternatively, take the cosine, and by the addition formula we get an identity,
$$\cos(\arccos x+\arcsin x)=\cos(\arccos x)\cos(\arcsin x)-\sin(\arccos x)\sin(\arcsin x)=x\sqrt{1-x^2}-\sqrt{1-x^2}x=0=\cos\left(\frac\pi2\right).$$
A: $$\sin^{-1}x=\theta $$
$$x=\sin\theta$$
$$x=\cos\left(\frac{\pi}{2}-\theta\right) $$
$$-\frac{\pi}{2} \le \theta \le \frac{\pi}{2}$$
$$ 0 \le \frac{\pi}{2} -\theta \le \pi$$
$$\frac{\pi}{2} -\theta=\cos^{-1}x$$
$$\frac{\pi}{2} -\sin^{-1}x=\cos^{-1}x$$
$$\sin^{-1}x + \cos^{-1}x=\frac{\pi}{2}$$
A: You started with an identity which means it is satisfied for all $x$. Stated in words what you said is 
"In a right triangle if the two acute angles sum up to $ \pi/2, $ what is one of them? Answer is of course "Any angle !"
If you had not a priori known the identity,then taking $cos$ of arguments on both sides of equation you get
$$ \cos(\, \sin^{-1} x + \cos^{-1} x )= \sqrt{1-x^2 }\, x -x \sqrt{1-x^2 }  =0  $$
So it can make your infer that angle can be any value. The same holds if on RHS $\pi/2 $ or $ 3 \pi/2 $ or those obtained by adding $ 2n\pi$ to them are given in the problem at start.
A: Here is a proof without words:

