# Trig: Basic Identities Problem

1. Mar 26, 2004

### Nik

Hi, today I was writing my Identities final test and got stuck on one question which I couldn't solve.

The question was: If $$sinB=-\frac{5}{13}$$ find the exact value of $$tan2B$$.

I've got this far, correct me if I went wrong anywhere: $$tan2B=\frac{sin2B}{cos2B}=\frac{2sinBcosB}{2cos^2B-1}$$

Then, when I tryied to solve further I came to a dead end, even though I tried many different ways.

2. Mar 26, 2004

### ShawnD

There should probably be 2 answers.

First of all, with that triangle, the hypotinuse is 13, x is +12 and -12, y is -5, and B is either in the third or fourth quadrant.

Fill in the equation using x as 12, then fill in the equation using -12.

Last edited: Mar 26, 2004
3. Mar 26, 2004

### Nik

Tnx for your help, the answers I've got are $$\frac{120}{119}$$ and $$\frac{-120}{119}$$

Once I get my test back I'll post if the answers match.

Last edited: Mar 27, 2004
4. Mar 27, 2004

### Muzza

If sin(B) = -5/13, then sin^2(B) = 25/169. Using the Pythagorean identity, we have that cos^2(B) = 1 - sin^2(B) = 1 - 25/169 = 144/169, which gives cos(B) = +/- 12/13. Plug those values into the formula for tan(2B) and voila...