Easy question Find $\sin 2x$, $\cos 2x$, and $\tan 2x$ Ok so I was absent from school yesterday because long story short I had no way to get to class b/c something happened last minute. I'm pretty sure this is easy but I keep getting the wrong answer for $\tan2x$.
Find $\sin 2x$, $\cos 2x$, and $\tan 2x$ from the given information.
$\tan x = -\frac{4}{3}$,   $x$ in Quadrant II
So I used the double angle formulas and got 
$\sin2x = -24/25$
$\cos2x = -7/25$
But I keep getting wrong answers for tangent (I got both $-200/27$ and $-24/7$ somehow). Can someone do a step by step guide on how to get $\tan2x$? Thank you.
 A: \begin{align}
\tan(2x) & = \frac{2\tan x}{1-\tan^2 x} = \frac{2(-4/3)}{1-(-4/3)^2} = \frac{-8/3}{1-16/9} = \frac{-24}{-7} = \frac{24} 7. \\[30pt]
\tan(2x) & = \frac{\sin(2x)}{\cos(2x)} = \frac{-24/25}{-7/25} = \frac{24} 7.
\end{align}
A: If one is having difficulty with a problem like this, then perhaps going back to the original principles would be prudent.
Consider the $\sin 2\theta$ problem

Using the sin law we have
$$\frac{\sin 2\theta}{2\sin\theta} = \frac{\sin \theta^C}{1}$$
Which reduces to
$$\sin 2\theta = 2\sin\theta\cos\theta$$
Using the law of cosines we have
$$(2\sin\theta)^2 =1^2+1^2-2\cdot1\cdot1\cdot\cos 2\theta$$
Which reduces to
$$\cos 2\theta = \cos^2\theta-\sin^2\theta$$
To find the $\tan$ identity, just divide
$$\tan 2\theta =\frac{\sin 2\theta}{\cos 2\theta}= \frac{2\sin\theta\cos\theta}{\cos^2\theta-\sin^2\theta}=\frac{\frac{2\sin\theta\cos\theta}{\cos^2\theta}}{\frac{\cos^2\theta-\sin^2\theta}{\cos^2\theta}}=\frac{2\tan\theta}{1-\tan^2\theta}$$
There is no need to memorize double angle formulas; just draw a few triangles. Math suddenly becomes a lot of fun. :)
