$(\sin^{-1} x)+ (\cos^{-1} x)^3$

Question

How do I find the least and maximum value of $(\sin^{-1} x)+ (\cos^{-1} x)^3$ ? I have tried the formula $(a+b)^3=a^3 + b^3 +3ab(a+b)$ , but seem to reach nowhere near ?

Answer 1

If $a+b=k,$

Method $\#1:$ $$a^3+b^3=(a+b)^3-3ab(a+b)=k^3-3kab$$

Now $$(a+b)^2-4ab=(a-b)^2\ge0\iff-4ab\ge-(a+b)^2$$

Method $\#2:$

$$a^3+b^3=a^3+(k-a)^3=k^3-3k^2a+3ka^2=k^3+3k\left(a^2-ka\right)$$

Now $a^2-ka=\dfrac{(2a-k)^2-k^2}4\ge-\dfrac{k^2}4$

For both methods, here $$k=\dfrac\pi2$$

Answer 2

$$\frac{d}{dx}\left(\arcsin x +\arccos^{3}x\right)$$ $$\frac{d}{dx}\left(\arcsin x\right) +\frac{d}{dx}\left(\arccos^{3}x\right)$$ $$\frac{1}{\cos(\arcsin x)}+3\arccos^2x\cdot\left(-\frac{1}{\sin(\arccos x)}\right)$$ $$\frac{1}{\sqrt{1-x^2}}+3\arccos^2x\cdot\left(-\frac{1}{\sqrt{1-x^2}}\right)$$ $$\frac{1-3\arccos^2 x}{\sqrt{1-x^2}}$$

This expression is equal to zero when $x=\cos\left(\frac{\sqrt{3}}{3}\right)$.

Plugging back into the original expression, we get a minimum value of $\frac{\pi}{2}-\frac{\sqrt{3}}{3}+\frac{\sqrt{3}}{9}$

Since there is no other place where the derivative equals zero, we conclude that the maximum must be at one of the two "endpoints" of the function. Trying both $x=1$ and $x=-1$, we see that the maximum occurs at $x=-1$, where $\left(\arcsin x +\arccos^{3}x\right)=\pi^3-\frac{\pi}{2}$

P.S. if you meant $\left(\arcsin x + \arccos x\right)^3$, that function is constant at $\frac{\pi^3}{8}$