# Math Help - Another exact value problem

1. ## Another exact value problem

how would you go about solving this:

if sin x = -4/5, and 270 degrees < x < 360 degrees, then find tan2x

2. Originally Posted by Kesha
how would you go about solving this:

if sin x = -4/5, and 270 degrees < x < 360 degrees, then find tan2x
$\tan (2x) = \frac{2 \tan x}{1 - \tan^2 x}$

From $\sin x = - \frac{4}{5}$ and the fact that x is in the fourth quadrant it follows that $\tan x = - \frac{4}{3}$.

Therefore .....

3. $\displaystyle\tan 2x=\frac{2\tan x}{1-\tan^2x}$ (1)
If $x\in (270^{\circ},360^{\circ})\Rightarrow \cos x>0$.
Then, $\cos x=\sqrt{1-\sin^2x}=\frac{3}{5}\Rightarrow\tan x=\frac{\sin x}{\cos x}=-\frac{4}{3}$.
Now, substitute $\tan x$ in (1) and find $\tan 2x$.

4. thanks, was easy to do. i just approached it incorrectly.