Trig: Basic Identities Problem

[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.

Thanx in advance!

[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.

[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.

[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...