1. Tan A

If sin A/2 = -12/13

Find Tan A.

I'm really clueless here. The A/2 made it complicated for me that I couldn't get sin A. Help please.

2. Originally Posted by reiward
If sin A/2 = -12/13

Find Tan A.

I'm really clueless here. The A/2 made it complicated for me that I couldn't get sin A. Help please.
HI

$\displaystyle \tan A=\frac{2\tan \frac{A}{2}}{1-\tan^2 \frac{A}{2}}$

$\displaystyle =\frac{2(\frac{\sin \frac{A}{2}}{\cos \frac{A}{2}})}{1-(\frac{\sin \frac{A}{2}}{\cos \frac{A}{2}})^2}$

3. Originally Posted by reiward
If sin A/2 = -12/13

Find Tan A.

I'm really clueless here. The A/2 made it complicated for me that I couldn't get sin A. Help please.
Since sin is negative, so it can be in quadrant 3 and 4 . Are you sure that the question did not mention bout cos A/2 or tan A/2 ? Or else you would get 2 different answers for tan A .

Since sin is negative, so it can be in quadrant 3 and 4 . Are you sure that the question did not mention bout cos A/2 or tan A/2 ? Or else you would get 2 different answers for tan A .
Oh, it mentioned that A/2 is a negative obtuse angle. So third quadrant?

Uhm, why did you use Tan 2A identities

5. Originally Posted by reiward
Oh, it mentioned that A/2 is a negative obtuse angle. So third quadrant?

Uhm, why did you use Tan 2A identities
Because it wouldn't lead me to the answer . I used the half angles identities instead .

Yes , A/2 would be in the 3rd quadrant .

so sin A/2 = .... and cos A/2 = .. , then substitute into the identity i gave you just now .