Answer
$\sin 2A=\frac{3}{5}$
$\sin \frac{A}{2}=\sqrt {\frac{10-\sqrt {10}}{20}}$
Work Step by Step
$\sec A=\frac{1}{\cos A}=\sqrt {10}$
$\implies\cos A=\frac{1}{\sqrt {10}}\cdot\frac{\sqrt {10}}{\sqrt {10}}=\frac{\sqrt {10}}{10}$
$\sin A=\pm\sqrt {1-\cos^{2}A}=\pm\sqrt {1-\frac{1}{10}}=\pm\frac{3\sqrt {10}}{10}$
$\sin A$ is positive as $A$ is in the first quadrant.
That is, $\sin A=\frac{3\sqrt {10}}{10}$
$\sin 2A=2\sin A\cos A=2\times\frac{3\sqrt {10}}{10}\times\frac{\sqrt {10}}{10}=\frac{3}{5}$
$\sin \frac{A}{2}=\pm\sqrt {\frac{1-\cos A}{2}}=\pm\sqrt {\frac{1-\frac{\sqrt {10}}{10}}{2}}=\pm\sqrt {\frac{10-\sqrt {10}}{20}}$
$\sin \frac{A}{2}$ is positive as $\frac{A}{2}$ terminates in the first quadrant (between $0^{\circ}$ and $45^{\circ}$).
So, $\sin \frac{A}{2}=\sqrt {\frac{10-\sqrt {10}}{20}}$
