## Trigonometry (11th Edition) Clone

$$m=\frac{2}{\sqrt{2-\sqrt3}}$$
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: $$\sqrt{\frac{1-\cos30^\circ}{2}}=\frac{1}{m}$$ $$\sqrt{\frac{1-\frac{\sqrt3}{2}}{2}}=\frac{1}{m}$$ $$\sqrt{\frac{\frac{2-\sqrt3}{2}}{2}}=\frac{1}{m}$$ $$\sqrt{\frac{2-\sqrt3}{4}}=\frac{1}{m}$$ $$\frac{\sqrt{2-\sqrt3}}{2}=\frac{1}{m}$$ $$m=\frac{2}{\sqrt{2-\sqrt3}}$$