# Given Csc(x)=8; how do you find sin x/2, cos x/2, tan x/2?

Jan 2, 2017

Two solutions - (A) if $\cos \left(\frac{x}{2}\right) = \frac{1}{2} \left(3 \sqrt{2} + \sqrt{14}\right)$, $\sin \left(\frac{x}{2}\right) = \frac{1}{2} \left(3 \sqrt{2} - \sqrt{14}\right)$ and $\tan \left(\frac{x}{2}\right) = 8 - 3 \sqrt{7}$

and (B) if $\cos \left(\frac{x}{2}\right) = \frac{1}{2} \left(3 \sqrt{2} - \sqrt{14}\right)$, $\sin \left(\frac{x}{2}\right) = \frac{1}{2} \left(3 \sqrt{2} + \sqrt{14}\right)$ and $\tan \left(\frac{x}{2}\right) = 8 + 3 \sqrt{7}$

#### Explanation:

As $\csc x = 8$, $\sin x = \frac{1}{\csc} x = \frac{1}{8}$ and as $\sin x > 0$, we have $0 < x < \pi$ and

$0 < \frac{x}{2} < \frac{\pi}{2}$ and hence $\frac{x}{2}$ lies on Q1 and all trigonometric ratios are positive.

As $\sin x = \frac{1}{8}$,

$\cos x = \sqrt{1 - {\left(\frac{1}{8}\right)}^{2}} = \sqrt{1 - \frac{1}{64}} = \pm \frac{\sqrt{63}}{8}$

and as $\cos 2 A = 2 {\cos}^{2} A - 1 = 1 - 2 {\sin}^{2} A$, we have

$\cos A = \sqrt{\frac{1 + \cos 2 A}{2}}$ and $\sin A = \sqrt{\frac{1 - \cos 2 A}{2}}$

Hence $\cos \left(\frac{x}{2}\right) = \sqrt{\frac{1 + \cos x}{2}}$ i.e.

$\frac{\sqrt{\left(1 \pm \frac{\sqrt{63}}{8}\right)}}{2} = \sqrt{\frac{8 \pm \sqrt{63}}{16}} = \frac{\sqrt{8 \pm 3 \sqrt{7}}}{4} = \frac{1}{2} \left(3 \sqrt{2} \pm \sqrt{14}\right)$

and $\sin \left(\frac{x}{2}\right) = \sqrt{\frac{1 - \cos x}{2}}$ i.e.

sqrt((1+-sqrt63/8))/2=sqrt((8+-sqrt63)/16)=sqrt(8+-3sqrt7)/4=1/2(3sqrt2∓sqrt14)

Note that (A) if $\cos \left(\frac{x}{2}\right) = \frac{1}{2} \left(3 \sqrt{2} + \sqrt{14}\right)$, $\sin \left(\frac{x}{2}\right) = \frac{1}{2} \left(3 \sqrt{2} - \sqrt{14}\right)$ and $\tan \left(\frac{x}{2}\right) = \frac{\sqrt{8 - \sqrt{63}}}{\sqrt{8 + \sqrt{63}}} = 8 - 3 \sqrt{7}$

and (B) if $\cos \left(\frac{x}{2}\right) = \frac{1}{2} \left(3 \sqrt{2} - \sqrt{14}\right)$, $\sin \left(\frac{x}{2}\right) = \frac{1}{2} \left(3 \sqrt{2} + \sqrt{14}\right)$ and $\tan \left(\frac{x}{2}\right) = \frac{\sqrt{8 + \sqrt{63}}}{\sqrt{8 - \sqrt{63}}} = 8 + 3 \sqrt{7}$