# How do you find cot 2B, given sin B = 12/13 and cos B < 0?

Nov 28, 2015

Find cot 2B, given $\sin B = \frac{12}{13}$ and cos B < 0

Ans: $\frac{119}{120}$

#### Explanation:

$\sin B = \frac{12}{13}$.
Find cos B by the identity: ${\cos}^{2} a + {\sin}^{2} a = 1$
${\cos}^{2} B = 1 - {\sin}^{2} B = 1 - \frac{144}{169} = \frac{25}{169}$ --> $\cos B = \pm \frac{5}{13.}$
$\cos B = - \frac{5}{13}$ (since cos B < 0)
Find: $\sin 2 B = 2 \sin B . \cos B = 2 \left(\frac{12}{13}\right) \left(- \frac{5}{13}\right) = - \frac{120}{169}$
Find: $\cos 2 B = 2 {\cos}^{2} B - 1 = 2 \left(\frac{25}{169}\right) - 1 = - \frac{119}{169.}$
Therefor: $\cot 2 B = \cos \frac{2 B}{\sin 2 B} = - \frac{119}{169} \left(\frac{169}{-} 120\right) = \frac{119}{120}$