# Double-Angle Identity

• Mar 18th 2009, 05:06 PM
lightningstab714
Double-Angle Identity
Find $\displaystyle \sin \left( {\theta} \right)$, $\displaystyle \cos \left( {\theta} \right)$, and $\displaystyle \tan \left( {\theta} \right)$ given that $\displaystyle \cot \left( {2\theta} \right) = -5/12$; 2θ is in Quadrant II.

I'm pretty sure solving this involves the double-angle (or maybe a half-angle) identity for tangent (the inverse of cotangent), but I'm not sure how to get $\displaystyle \cot \left( {2\theta} \right) = -5/12$ in terms of tangent. Please just help me get this started. I can solve for $\displaystyle \sin \left( {\theta} \right)$ and $\displaystyle \cos \left( {\theta} \right)$ if it were written in terms in terms of tangent. Thanks.
• Mar 18th 2009, 06:13 PM
skeeter
Quote:

Originally Posted by lightningstab714
Find $\displaystyle \sin \left( {\theta} \right)$, $\displaystyle \cos \left( {\theta} \right)$, and $\displaystyle \tan \left( {\theta} \right)$ given that $\displaystyle \cot \left( {2\theta} \right) = -5/12$; 2θ is in Quadrant II.

I'm pretty sure solving this involves the double-angle (or maybe a half-angle) identity for tangent (the inverse of cotangent), but I'm not sure how to get $\displaystyle \cot \left( {2\theta} \right) = -5/12$ in terms of tangent. Please just help me get this started. I can solve for $\displaystyle \sin \left( {\theta} \right)$ and $\displaystyle \cos \left( {\theta} \right)$ if it were written in terms in terms of tangent. Thanks.

$\displaystyle \cot(2t) = -\frac{5}{12}$

$\displaystyle \tan(2t) = -\frac{12}{5}$

$\displaystyle \tan(2t) = \frac{2\tan{t}}{1 - \tan^2{t}}$

$\displaystyle -\frac{12}{5} = \frac{2\tan{t}}{1 - \tan^2{t}}$

$\displaystyle 6\tan^2{t} - 5\tan{t} - 6 = 0$

$\displaystyle (3\tan{t} + 2)(2\tan{t} - 3) = 0$

since $\displaystyle \frac{\pi}{2} < 2t < \pi$

$\displaystyle \frac{\pi}{4} < t < \frac{\pi}{2}$

$\displaystyle \tan{t} = \frac{3}{2}$