<s>[INST] 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?
 [/INST] 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 $f$:

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