How to find the range of $ (\arcsin x)^2 + (\arccos x)^2 $ without using derivatives?

Question

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?

Answer 1

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 $-\dfrac\pi2\le y\le\dfrac\pi2,-\dfrac\pi2-\dfrac\pi4\le y-\dfrac\pi4\le\dfrac\pi2-\dfrac\pi4$

$\implies0\le\left(y-\dfrac\pi4\right)^2\le\left(-\dfrac\pi2-\dfrac\pi4\right)^2$

Answer 2

Adding the ranges is wrong: what about $g(x)=x^2+x$?

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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 $$ \left[\frac{\tau^2}{2},5\tau^2\right]=\left[\frac{\pi^2}{8},\frac{5\pi^2}{4}\right] $$

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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 $x=1$.

Answer 3

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 $f$.

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