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Given $x,y$ are acute angles such that $$\sin y = 3 \cos(x+y)\sin x$$ Find the maximum value of $\tan ⁡y$.

Attempt: We have

$$\begin{aligned} 3(\cos x \cos y - \sin x \sin y) \sin x & = \sin y \\ 3 \cos x \sin x - 3 \sin^2 x \tan y & = \tan y \\ 3 \cos x \sin x & = \tan y(1 +3 \sin^2 x) \\ \tan y & = \dfrac{3 \sin x \cos x} {1+3 \sin^2 x} \end{aligned}$$

Now, how about the next step? Or maybe I did some mistakes?

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    $\begingroup$ I think now you could use $1 = \sin^2 x + \cos^2 x$, then divide the top and the bottom by $\cos^2 x$ and transform the equation to a fraction of $\tan x$. $\endgroup$
    – xbh
    Jan 17, 2019 at 6:49

3 Answers 3

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By AM-GM $$\tan{y}=\frac{3\sin{x}\cos{x}}{1+3\sin^2x}=\frac{3\sin{x}\cos{x}}{\cos^2x+4\sin^2x}\leq\frac{3\sin{x}\cos{x}}{2\sqrt{\cos^2x\cdot4\sin^2x}}=\frac{3}{4}.$$ The equality occurs for $\cos{x}=2\sin{x},$ which says that we got a maximal value.

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  • $\begingroup$ Thank you very much, Sir. $\endgroup$ Jan 17, 2019 at 7:09
  • $\begingroup$ @Shane Dizzy Sukardy You are welcome! $\endgroup$ Jan 17, 2019 at 7:10
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Put $t = \tan x$, then $\tan y = \dfrac{3t}{1+4t^2}\le \dfrac{3t}{4t} = \dfrac{3}{4}$, which is the max of $\tan y$ with equality occurs when $t = \dfrac{1}{2}$ or $2\sin x = \cos x$...the rest is simple...

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$$\tan y=\dfrac{3\tan x}{1+4\tan^2x}$$

$$\iff(4\tan y)\tan ^2x-3\tan x+\tan y=0$$ which is a Quadratic Equation $\tan x$

As $\tan x$ is real, the discriminant must be $\ge0$

i.e., $$3^2-4(4\tan y)\ge0\tan y\iff\tan^2y\le\dfrac9{16}\iff-\dfrac34\le\tan y\le\dfrac34$$

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