Answer
$$m=\frac{2}{\sqrt{2-\sqrt3}}$$
Work Step by Step
Formula: $$\sin\frac{\theta}{2}=\frac{1}{m}$$
$$\theta=30^\circ$$
- From half-angle identity: $$\sin\frac{\theta}{2}=\pm\sqrt{\frac{1-\cos\theta}{2}}$$
Replace the angle into the identity:
$$\sin\frac{30^\circ}{2}=\pm\sqrt{\frac{1-\cos30^\circ}{2}}$$
$$\sin15^\circ=\pm\sqrt{\frac{1-\cos30^\circ}{2}}$$
$15^\circ$ is in quadrant I, where sines are positive. Therefore, $\sin15^\circ\gt0$ and we need to pick the positive square root as a result.
$$\sin15^\circ=\sqrt{\frac{1-\cos30^\circ}{2}}$$
Combining the identity with the formula:
$$\sqrt{\frac{1-\cos30^\circ}{2}}=\frac{1}{m}$$
$$\sqrt{\frac{1-\frac{\sqrt3}{2}}{2}}=\frac{1}{m}$$
$$\sqrt{\frac{\frac{2-\sqrt3}{2}}{2}}=\frac{1}{m}$$
$$\sqrt{\frac{2-\sqrt3}{4}}=\frac{1}{m}$$
$$\frac{\sqrt{2-\sqrt3}}{2}=\frac{1}{m}$$
$$m=\frac{2}{\sqrt{2-\sqrt3}}$$
