<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] Your strategy is correct, but you failed to take into account that $x \in \left[\pi, \frac{3\pi}{2}\right] \implies \sin x < 0$.
As for an alternative method:  
We use the identity 
$$\sin(3x) = 3\cos^2x\sin x - \sin^3x$$ 
Using the identity $\sec^2x = \tan^2x + 1$ yields
$$\sec^2x = (2\sqrt{2})^2 + 1 = 8 + 1 = 9 \implies \cos^2x = \frac{1}{\sec^2x} = \frac{1}{9}$$
Using the Pythagorean identity $\sin^2x + \cos^2x = 1$ yields 
$$\sin^2x = 1 - \cos^2x$$
Since $x \in \left[\pi, \frac{3\pi}{2}\right]$, $\sin x < 0$.  Hence,
$$\sin x = -\sqrt{1 - \cos^2x} = -\sqrt{1 - \frac{1}{9}} = -\sqrt{\frac{8}{9}} = -\frac{2\sqrt{2}}{3}$$
Substitution into the identity $\sin(3x) = 3\cos^2x\sin x - \sin^3x$ yields
$$\sin(3x) = 3\left(\frac{1}{9}\right)\left(-\frac{2\sqrt{2}}{3}\right) - \left(-\frac{2\sqrt{2}}{3}\right)^3 = -\frac{6\sqrt{2}}{27} + \frac{16\sqrt{2}}{27} = \frac{10\sqrt{2}}{27}$$
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