# Given sin(alpha)=4/9 and pi/2<alpha<pi, find the exact value of sin(alpha/2)

Given $\mathrm{sin}\left(\alpha \right)=\frac{4}{9}\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\pi \text{/}2<\alpha <\pi$,
find the exact value of $\mathrm{sin}\left(\alpha \text{/}2\right).$

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Step 1
$\frac{\pi }{2}<\alpha <\pi$
$\frac{\frac{\pi }{2}}{2}<\frac{\alpha }{2}<\frac{\pi }{2}$
$\frac{\pi }{4}<\frac{\alpha }{2}<\frac{\pi }{2}$
So, $\frac{\alpha }{2}$ in forst quadrant.
In first quadrant sin is positive.
Step 2 $\mathrm{sin}\left(\frac{\alpha }{2}\right)=\sqrt{\frac{1-\mathrm{cos}\alpha }{2}}$
So, we have to find cos alpha from $\mathrm{sin}\alpha$
$\frac{\pi }{2}<\alpha <\pi$ is second quadrant.
There $\mathrm{cos}\alpha$ is negative
$\mathrm{cos}\alpha =-\sqrt{1-{\mathrm{sin}}^{2}\alpha }=-\sqrt{1-{\left(\frac{4}{9}\right)}^{2}}=-\sqrt{1-\frac{16}{81}}=-\frac{\sqrt{65}}{9}$
Step 3
Plug $\mathrm{cos}\alpha =-\frac{\sqrt{65}}{9}$ in the formula
$\mathrm{sin}\left(\frac{\alpha }{2}\right)=\sqrt{\frac{1-\mathrm{cos}\alpha }{2}}$
$\mathrm{sin}\left(\frac{\alpha }{2}\right)$
$=\sqrt{\frac{1-\left(-\frac{\sqrt{65}}{9}\right)}{2}}$
$=\sqrt{\frac{1+\frac{\sqrt{65}}{9}}{2}}$
$=\sqrt{\frac{9+\sqrt{65}}{18}}$
Multiply numerator and denominator by 2
$=\sqrt{\frac{2\left(9+\sqrt{65}\right)}{18\left(2\right)}}$
$=\sqrt{\frac{18+2\sqrt{65}}{36}}$
$=\sqrt{13+2\sqrt{13}\frac{\sqrt{5+5}}{6}}$
$=\frac{\sqrt{{\left(\sqrt{13}+\sqrt{5}\right)}^{2}}}{6}$
$=\frac{\sqrt{13}+\sqrt{5}}{6}$
Answer: $=\frac{\sqrt{13}+\sqrt{5}}{6}$