How do you find the exact value of cos (theta/2), given that csc theta = - 3/2 and tan theta > 0?

Jan 2, 2017

$\cos \left(\frac{\theta}{2}\right) = - \frac{\sqrt{15} - \sqrt{3}}{6}$

Explanation:

As $\csc \theta = - \frac{3}{2}$ and $\tan \theta > 0$

we have $\sin \theta = \frac{1}{- \frac{3}{2}} = - \frac{2}{3}$

As sine ratio is negative and tangent ratio is positive, $\theta$ lies in Q3 and hence $\cos \theta$ will be negative and as $\pi < \theta < \frac{3 \pi}{2}$, we have

$\frac{\pi}{2} < \frac{\theta}{2} < \frac{3 \pi}{4}$ and $\cos \left(\frac{\theta}{2}\right)$ is too negative.

and $\cos \theta = \sqrt{1 - {\left(- \frac{2}{3}\right)}^{2}} = \sqrt{1 - \frac{4}{9}} = - \frac{\sqrt{5}}{9} = - \frac{\sqrt{5}}{3}$

and as $\cos \theta = 2 {\cos}^{2} \left(\frac{\theta}{2}\right) - 1$

$\cos \left(\frac{\theta}{2}\right) = \sqrt{\frac{1 + \cos \theta}{2}} = = \sqrt{\frac{1 - \frac{\sqrt{5}}{3}}{2}}$

= $- \sqrt{\frac{3 - \sqrt{5}}{6}} = - \sqrt{\frac{18 - 6 \sqrt{5}}{36}}$

= $- \sqrt{\frac{15 + 3 - 2 \sqrt{45}}{36}}$

= $- \sqrt{\frac{{\left(\sqrt{15}\right)}^{2} + {\left(\sqrt{3}\right)}^{2} - 2 \sqrt{15 \times 3}}{36}}$

= $- \frac{\sqrt{15} - \sqrt{3}}{6}$