How to evaluate inverse trigonometric functions in exponents? I recently started working on inverse trigonometry, I have done many problems that include conversion of inverse functions and many more formulas but how should we approach a question when the inverse functions are given in exponents.
For example, I came to a question,

$12^{\arcsin(x)} + 12^{\arccos(x)} + 12^{\arctan(x)} >3\cdot k^{\pi/k}$
Find $k$?

Can you please provide me a start. I will solve the rest on my own.
Thanks.
 A: I know no general method to approach a question when the inverse functions are given in exponents, but I guess that in general such questions hide simple ideas under a cumbersome form with artificially chosen functions and coefficients.. For instance, in this question the inequality holds for all positive $k$. We can show this as follows. 
Since $\operatorname{arcsin} x+\operatorname{arccos} x=\pi/2$ for each $x$, 
$$12^{\operatorname{arcsin} x}+12^{\operatorname{arccos} x}+12^{\operatorname{arctan} x}>$$ 
$$12^{\operatorname{arcsin} x}+12^{\operatorname{arccos} x}\ge \mbox{ (by AM-GM) }$$ $$
12^{(\operatorname{arcsin} x+\operatorname{arccos} x)/2}=2\cdot 12^{\pi/4}\approx 14.080.$$
On the other hand, the derivative of the function $3k^{\pi/k}$ is $\frac{3\pi k^{\pi/k}}{k^2}(1-\ln k)$. Therefore the function increases when $0<k<e$, attains its maximum $3e^{\pi/e}\approx 9.529$ at $k=e$, and decreases when $k>e$. 
PS. The following graph (drawn by Mathcad) suggests that the minimum value of a function
$f(x)=12^{\operatorname{arcsin} x}+12^{\operatorname{arccos} x}+12^{\operatorname{arctan} x}$ 
is about $18.5237$ and is attained when $x$ is a solution of a transcendental equation $f’(x)=0$, that is when
$$(12^{\operatorname{arccos} x}-12^{\operatorname{arcsin} x})(1+x^2)=12^{\operatorname{arctan} x}\sqrt{1-x^2}.$$

A: Note that by AM-GM:
$$\frac{12^{\arcsin x} + 12^{\arccos x} + 12^{\arctan x}}3 \geq \sqrt[3]{12^{\arcsin x+\arccos x+\arctan x}}$$
and the following trigonometric identities hold:
$$\arcsin x+\arccos x = \frac{\pi}2$$
$$\arctan x +\arctan \frac1x = \begin{cases}\frac{\pi}2 \quad x>0\\-\frac{\pi}2 \quad x<0\end{cases}$$
thus:
$$\arcsin x+\arccos x+\arctan x=\begin{cases}\frac{\pi}2 + \arctan x\quad x>0\\-\arctan \frac1x\quad x<0\end{cases}$$
therefore
$$12^{\arcsin x} + 12^{\arccos x} + 12^{\arctan x} \geq \begin{cases}
3\sqrt[3]{12^{\frac{\pi}2 + \arctan x}}\quad x\in[0,1]\\
3\sqrt[3]{12^{-\arctan \frac1x}}\quad x\in[-1,0)
\end{cases}$$
$3\sqrt[3]{12^{\frac{\pi}2 + \arctan x}}$ plot
$3\sqrt[3]{12^{-\arctan \frac1x}}$ plot
A: You can try also to convert arcsin and arccos to their complex log expressions. If do that, the x goes down, out of the exponents.
