why $\sin^{-1}(x)+\cos^{-1}(x) = π/2$ How do i  find  the  When $x$ is  ranging from $-1$ to $1$?
I want know why $\sin^{-1}(x)+\cos^{-1}(x) = π/2$
I have already tried inverse-function.
thanks.
 A: There are a couple of ways of doing this; here's a short, simple geometric way. First, we draw a right triangle and label the vertices $A,B,C$ and the side lengths $a,b,c$.

We have$$\begin{align*} & \sin\theta=\frac bc\implies\theta=\arcsin\frac bc\\ & \cos\varphi=\frac bc\implies\varphi=\arccos\frac bc\end{align*}$$Obviously, the sum of $\theta$ and $\varphi$ is $90^{\circ}$. Therefore, it follows that$$\arcsin\frac bc+\arccos\frac bc=\frac {\pi}2$$
A: $$\frac{d}{dx}(\arccos x+\arcsin x)$$
$$=-\frac{1}{\sqrt{1-x^2}}+\frac{1}{\sqrt{1-x^2}}=0$$
So $\arccos x+\arcsin x$ is constant and letting $x=0$ reveals what that constant is. 

As a more elementary method if $\sin a=x$ with $a \in [-\frac{\pi}{2},\frac{\pi}{2}]$ and $\cos b=x$ with $b \in [0,\pi]$ then $\sin (\frac{\pi}{2}-b)=x$ and so $a=\frac{\pi}{2}-b$ follows by one-to-oneness on the intervals.
$$a+b=\arcsin x+\arccos x=\frac{\pi}{2}$$
A: The "co" in "cosine" stands for "complement".  It means, the sine of the complementary angle.  Two angles are complementary if they add to a right angle, $\pi/2$ radians.  Thus:
$$
\arcsin x + \arccos x = \frac{\pi}{2}
$$
A: Draw a right triangle whose hypotenuse has length $1$ and say the side of it opposite one of the angles, $\theta$ has length $x.$ Then the side of it adjacent to the other acute angle is that same side of length $x$. The other acute angle is $\pi/2-\theta.$
So $\theta = \sin^{-1} x$ and $\pi/2-\theta = \cos^{-1} x.$ Add those together.
