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

Given $$\displaystyle \sin{{\left(\alpha\right)}}=\frac{4}{{9}}{\quad\text{and}\quad}\pi\text{/}{2}<\alpha<\pi$$,
find the exact value of $$\displaystyle \sin{{\left(\alpha\text{/}{2}\right)}}.$$

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davonliefI

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