<s>[INST] Given $\tan(x) = 2\sqrt2 $ , $ x\in[ \pi , \frac{3\pi}{2}] $ , What is the exact value of $\sin(3x)$? 

Given $\tan(x) = 2\sqrt2 $ , $ x\in[ \pi , \frac{3\pi}{2}] $
What is the exact value of $\sin(3x)$?



What I have done: 
Given $\tan(x) = 2\sqrt2 $ , I drew a right angled triangle and found the hypotenuse to be $3$ so $\sin(x) = \frac{2\sqrt2}{3}$
Recalling that $$\sin(3x) = 3\sin(x) - 4\sin^3(x)$$
Subbing in $\sin(x) = \frac{2\sqrt2}{3}$ and $\sin^3(x) = \frac{16\sqrt2}{27}$
$$\sin(3x) = 3(\frac{2\sqrt2}{3}) - 4(\frac{16\sqrt2}{27})$$
$$ \sin(3x) = {2\sqrt2} - \frac{64\sqrt2}{27} $$
$$ \sin(3x) = \frac{-10\sqrt2}{27} $$
Is this correct? If so would there be any other way I could have gone about solving this?
 [/INST] $$\sin^2x=\frac1{1+\cot^2x}=\frac1{1+\frac1{8}}=\frac89\implies \sin x=-\frac{2\sqrt2}3\,,\,\,\text{since}\;\;x\in[\pi,\,3\pi/2]\implies$$
$$\sin3x=3\sin x-4\sin^3x=-2\sqrt2+\frac{64\sqrt2}{27}=\frac{10\sqrt2}{27}$$
since
$$\pi\le x\le\frac{3\pi}2\implies3\pi\le 3x\le4\pi+\frac\pi2$$
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