Another exact value problem

**Kesha** · July 2nd 2008, 09:30 PM

how would you go about solving this:

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

**mr fantastic** · July 2nd 2008, 10:27 PM

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 .....

**red_dog** · July 2nd 2008, 10:31 PM

$\displaystyle\tan 2x=\frac{2\tan x}{1-\tan^2x}$ (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 $\tan 2x$ .

**Kesha** · July 2nd 2008, 11:52 PM

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

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