prove the following trig identity For any $x \in [0,1]$ show that $$\arcsin(x)+\arccos(x)=\frac{\pi}{2}$$ Please note that this is not a homework problem, this is something I came across that appears to be true.
 A: Let $\displaystyle\arcsin x=\phi\implies$
$\displaystyle (i)x=\sin\phi $ 
and $\displaystyle (ii)-\frac\pi2\le \phi\le\frac\pi2$ based on the definition of principal value of inverse sine function
Now, $\displaystyle x=\sin\phi=\cos\left(\frac\pi2-\phi\right)$ the angle complies with the principal value of inverse cosine function $[0, \pi]$
$\displaystyle\implies \arccos x=\frac\pi2-\phi$
Remember this identity holds true for $x\in[-1,1]$
A: Let 
$$f(x)=\arcsin(x)+\arccos(x)$$
and since we have $$f'(x)=\frac{1}{\sqrt{1-x^2}}-\frac{1}{\sqrt{1-x^2}}=0$$
then
$$f(x)=f(0)=\frac \pi 2$$
Edit:  I assume that the expressions of the derivative of $\arcsin$ and $\arccos$  functions are obtained using the derivative of inverse function.
Second edit: To clarify my answer after some comments: We know that the function $f$ is differentaible only on the interval $[0,1)$ so $f$ is a constant equal to $\frac \pi 2$ on this interval but we can deduce the desired result since $f$ is continuous on the interval $[0,1]$.
A: Draw a right triangle.  Let $x$ be the sine of one of the (non-right) angles.  Then it's the cosine of the other.  If you know that the three angles must add up to $180^\circ$ and one of the three is $90^\circ$, then that does it.
A: Setting $$\arcsin(x)= a,$$ we get $$\sin(a)= x.$$
Also, we have $$\sin(a)=\cos(π/2-a)=x.$$
So we have $$\arccos (x)=\pi/2-a.$$
A: As $x\in[0,1],$ $\displaystyle0\le \arcsin x,\arccos x\le\frac\pi2$ 
If $\displaystyle\arcsin x=A,\arccos x=B,$
$\displaystyle \sin A=x\implies \cos A=+\sqrt{1-x^2},\cos B=x\implies\sin B=+\sqrt{1-x^2}$
$\displaystyle \sin(A+B)=\sin A\cos B+\sin B\cos A=x\cdot x+\sqrt{1-x^2}\cdot\sqrt{1-x^2}=1$
As $\displaystyle 0\le A,B\le\frac\pi2, 0\le A+B\le\pi, A+B=\frac\pi2$
