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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.

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  • $\begingroup$ Use AM-GM Inequality. Also bear in mind the admissible values for $x$ here are from the interval $[-1,1]$ $\endgroup$ – Hari Shankar May 17 '17 at 7:56
  • $\begingroup$ @HariShankar what should we use x as? $\endgroup$ – prog_SAHIL May 19 '17 at 3:47
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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

$$(12^{\operatorname{arccos} x}-12^{\operatorname{arcsin} x})(1+x^2)=12^{\operatorname{arctan} x}\sqrt{1-x^2}.$$

enter image description here

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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

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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.

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