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

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 ?

• Just to clarify, you want us to maximize/minimize $\sin^{-1}x+\left(\cos^{-1}x\right)^3$?
– Hrhm
Commented Jun 10, 2016 at 17:59
• Check the sign of the derivative... It has a unique zero, so extrema will be there or at the boundaries of the domain, $[-1,1]$. Commented Jun 10, 2016 at 18:04
• Do you mean $(\sin^{-1}x+\cos^{-1}x)^3$?
– Hrhm
Commented Jun 10, 2016 at 18:05
• lets do both $(\sin^{−1}x+\cos^{−1}x)^3$ and $\sin^{−1}⁡x+(\cos^{−1}⁡x)^3$ for him... ☺ Commented Jun 10, 2016 at 18:11

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

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 $\frac{\pi^3}{8}$

• There is certainly a maximum and it occurs at $-1$ for which the function is $\pi^3-\pi/2$. Commented Jun 10, 2016 at 19:03
• @Dr.MV Sorry, I overlooked that. Thank you for pointing that out, I made an edit. I have a quick question: Do you have a good explanation for why the maximum of this function must occur at either $x=1$ or $x=-1$?
– Hrhm
Commented Jun 10, 2016 at 19:11
• For a $C^1$ function, extrema is at end points or when the derivative is zero. Checking values then suffices. +1 Commented Jun 10, 2016 at 21:02

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 $$k=\dfrac\pi2$$

• And exactly how are those obvious facts related to the question? Commented Jun 11, 2016 at 4:41
• @Macavity, I still believe the question to be $$(\sin^{-1}x)^3+(\cos^{-1}x)^3$$ Now the relation should be obvious,right? Commented Jun 11, 2016 at 4:47
• Ok I see. As written currently that isn't the question, but your interpretation seems likely. +1 Commented Jun 11, 2016 at 5:43