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.

• Use your correct calculations of $\sin(2x)$ and $\cos(2x)$, and divide. Note that the minus signs "cancel." – André Nicolas Jul 31 '15 at 23:12
• Recall that $\tan\theta=\frac{\sin\theta}{\cos\theta}$, here $\theta=2x$ – user265675 Jul 31 '15 at 23:13
• Oh I was using the double angle formula for tangent lol, that way is much easier. Thank you! – TheNewGuy Jul 31 '15 at 23:17

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. :)