# How do you find sin(x/2) if cscx=3?

May 17, 2015

Since csc x=3, sin x= $\frac{1}{3}$ and therefore cos x=$\sqrt{\frac{8}{9}}$= $\frac{2 \sqrt{2}}{3}$

Since cosx equals $1 - 2 {\sin}^{2} \left(\frac{x}{2}\right)$, hence

$2 {\sin}^{2} \left(\frac{x}{2}\right) = 1 - \cos x$

= 1- $\frac{2 \sqrt{2}}{3}$ =$\frac{3 - 2 \sqrt{2}}{3}$ = ${\left(\sqrt{2} - 1\right)}^{2} / 3$

${\sin}^{2} \left(\frac{x}{2}\right) = {\left(\sqrt{2} - 1\right)}^{2} / 6$

$\sin \left(\frac{x}{2}\right) = \frac{\sqrt{2} - 1}{\sqrt{6}}$

May 17, 2015

Use trig identity:$1 + {\cot}^{2} x = \frac{1}{\sin} ^ 2 x = {\csc}^{2} x$

$\frac{1}{\sin} ^ 2 x = 9 \to {\sin}^{2} x = \frac{1}{9} \to \sin x = + \frac{1}{3}$

$\cos x = 1 - {\sin}^{2} x = 1 - \frac{1}{9} = \frac{8}{9} \to \cos x = + \frac{\sqrt{2}}{3}$

To find $\sin \left(\frac{x}{2}\right)$ use trig identity: $2 {\sin}^{2} \left(\frac{x}{2}\right) = 1 - \cos x .$

$2 {\sin}^{2} \left(\frac{x}{2}\right) = 1 - \cos x = \frac{3}{3} - \frac{\sqrt{2}}{3} = 0.53$

${\sin}^{2} \left(\frac{x}{2}\right) = 0.265 \to \sin \left(\frac{x}{2}\right) = 0.51$