## Trigonometry (11th Edition) Clone

$$\sin15^\circ=\frac{\sqrt{2-\sqrt3}}{2}$$ 5 comes with C.
$$\sin15^\circ$$ $15^\circ$ can be written as half of the angle $30^\circ$. $$15^\circ=\frac{30^\circ}{2}$$ Thus, $$\sin15^\circ=\sin\Big(\frac{30^\circ}{2}\Big)$$ We can apply here the Half-Angle Identity for sine, which states $$\sin\Big(\frac{A}{2}\Big)=\pm\sqrt{\frac{1-\cos A}{2}}$$ with $A=30^\circ$: $$\sin15^\circ=\pm\sqrt{\frac{1-\cos30^\circ}{2}}$$ $15^\circ$ is in quadrant I. As $\sin\theta\gt0$ as $\theta$ is in quadrant I, $\sin15^\circ\gt0$. So we would pick the positive square root. $$\sin15^\circ=\sqrt{\frac{1-\frac{\sqrt3}{2}}{2}}$$ $$\sin15^\circ=\sqrt{\frac{\frac{2-\sqrt3}{2}}{2}}$$ $$\sin15^\circ=\sqrt{\frac{2-\sqrt3}{4}}$$ $$\sin15^\circ=\frac{\sqrt{2-\sqrt3}}{2}$$ Therefore, 5 needs to be matched with C.