# If $x$ and $y$ are acute, and $\sin y = 3 \cos (x+y) \sin x$⁡, then find the maximum value of $\tan y$

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?

• 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 $\tan x$.
– xbh
Jan 17, 2019 at 6:49

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.

• Thank you very much, Sir. Jan 17, 2019 at 7:09
• @Shane Dizzy Sukardy You are welcome! Jan 17, 2019 at 7:10

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...

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

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

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

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