How to find the range of $ (\arcsin x)^2 + (\arccos x)^2 $ without using derivatives? I am expected to find the range of the following function:
$$
(\arcsin x)^2 + (\arccos x)^2
$$
I individually added up the range of $(\arcsin x)^2$ $(\arccos x)^2$
But that gave wrong answer ?
What to do with this problem?
 A: Let $\arcsin x=y$
$$(\arcsin x)^2+(\arccos x)^2=y^2+\left(\dfrac\pi2-y\right)^2$$
$$=2y^2+\dfrac{\pi^2}4-\pi y=2\left(y-\dfrac\pi4\right)^2+\dfrac{\pi^2}4-\dfrac{2\pi^2}{16}$$
$\left(y-\dfrac\pi4\right)^2\ge0$
Again $-\dfrac\pi2\le y\le\dfrac\pi2,-\dfrac\pi2-\dfrac\pi4\le y-\dfrac\pi4\le\dfrac\pi2-\dfrac\pi4$
$\implies0\le\left(y-\dfrac\pi4\right)^2\le\left(-\dfrac\pi2-\dfrac\pi4\right)^2$
A: Adding the ranges is wrong: what about $g(x)=x^2+x$?

Set $\tau=\pi/2$ for simplicity. Then $\arccos x=\tau-\arcsin x$ and your function is
$$
2\arcsin^2x-2\tau\arcsin x+\tau^2
$$
The graph of $f(X)=2X^2-2\tau X+\tau^2$ is a convex parabola, of which you have to consider only the part with $-\tau\le X\le\tau$.
The axis of the parabola is $X=\tau/2$, so the function is decreasing on $[-\tau,\tau/2]$ and increasing on $[\tau/2,\tau]$. Since
\begin{align}
F(-\tau)&=2\tau^2+2\tau^2+\tau^2=5\tau^2 \\[4px]
F(\tau/2)&=\frac{\tau^2}{2}-\tau^2+\tau^2=\frac{\tau^2}{2} \\[4px]
F(\tau)&=2\tau^2-2\tau^2+\tau^2=\tau^2,
\end{align}
the range is
$$
\left[\frac{\tau^2}{2},5\tau^2\right]=\left[\frac{\pi^2}{8},\frac{5\pi^2}{4}\right]
$$

With calculus, if $f(x)=2\arcsin^2x-2\tau\arcsin x+\tau^2$, then
$$
f'(x)=\frac{4\arcsin x}{\sqrt{1-x^2}}-\frac{2\tau}{\sqrt{1-x^2}}=
\frac{2(2\arcsin x-\tau)}{\sqrt{1-x^2}}
$$
The derivative vanishes where $\arcsin x=\tau/2=\pi/4$, a local minimum, actually absolute minimum. We find exactly the same points as before, because we also need to take care of the points where $f$ is not differentiable, that is at $x=-1$ and $x=1$.
A: Let me write $f(x) = \arcsin^2 x + \arccos^2 x$. The natural domain of $f$ is the intersection of natural domains of $\arcsin$ and $\arccos$: $$\mathcal D(f) = \mathcal D(\arcsin) \cap \mathcal D(\arccos) = [-1,1]\cap [-1,1] = [-1,1].$$
From the identity $\sin x = \cos(\frac\pi 2 - x)$, it follows that $\arcsin x + \arccos x = \frac\pi 2$. Thus, we have $$ f(x) = \arcsin^2 x + \arccos^2 x = \arcsin^2 x + (\frac\pi 2 - \arcsin x)^2 = 2\arcsin^2 x -\pi \arcsin x + \frac{\pi^2}4.$$
Let us define $g(x) = 2x^2 - \pi x + \frac{\pi^2}4$. We can see that $f(x) = g(\arcsin x)$. The range of $f$ is thus:
$$f([-1,1]) = g(\arcsin ([-1,1])) = g([-\frac\pi 2,\frac\pi 2]).$$
The graph of $g$ is a parabola. To determine $g([-\frac\pi 2,\frac\pi 2])$, let us first write $g(x) = 2(x-\frac\pi 4)^2+\frac{\pi^2}8$. 
The minimum of $g$ on $\mathbb R$ occurs at $x_0 = \frac\pi 4$ and is equal to $\frac{\pi^2}8$. Since $x_0\in [-\frac\pi 2,\frac\pi 2]$, it is also the minimum of $g$ on the segment $[-\frac\pi 2,\frac\pi 2]$. 
The maximum of $g$ then occurs at one of the boundary points $\{\pm\frac\pi 2\}$. Calculate $g(\frac\pi 2) = \frac{\pi^2}4$ and $g(-\frac\pi 2) = \frac{5\pi^2}4$. Since $g(-\frac\pi 2)$ is greater, that is the maximum of $g$ on $[-\frac\pi 2,\frac\pi 2]$.
We conclude that $g([-\frac\pi 2,\frac\pi 2]) = [\frac{\pi^2}8,\frac{5\pi^2}4]$ and that is the range of $f$.

Visual aid. The red line is the range of $\arcsin$ and the blue line is the range of $f$:

