1. ## Double-Angle Identity

Find $\sin \left( {\theta} \right)$, $\cos \left( {\theta} \right)$, and $\tan \left( {\theta} \right)$ given that $\cot \left( {2\theta} \right) = -5/12$; 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 $\cot \left( {2\theta} \right) = -5/12$ in terms of tangent. Please just help me get this started. I can solve for $\sin \left( {\theta} \right)$ and $\cos \left( {\theta} \right)$ if it were written in terms in terms of tangent. Thanks.

2. Originally Posted by lightningstab714
Find $\sin \left( {\theta} \right)$, $\cos \left( {\theta} \right)$, and $\tan \left( {\theta} \right)$ given that $\cot \left( {2\theta} \right) = -5/12$; 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 $\cot \left( {2\theta} \right) = -5/12$ in terms of tangent. Please just help me get this started. I can solve for $\sin \left( {\theta} \right)$ and $\cos \left( {\theta} \right)$ if it were written in terms in terms of tangent. Thanks.
$\cot(2t) = -\frac{5}{12}$

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

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

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

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

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

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

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

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