## Trigonometry (11th Edition) Clone

$$\sin7.5^\circ=\frac{\sqrt{2-\sqrt{2+\sqrt3}}}{2}$$
$$\sin7.5^\circ$$ As $7.5^\circ$ is half the angle $15^\circ$, $$\sin7.5^\circ=\sin\Big(\frac{15^\circ}{2}\Big)$$ Apply the following half-angle identity for sine for $A=15^\circ$: $$\sin\Big(\frac{A}{2}\Big)=\pm\sqrt{\frac{1-\cos A}{2}}$$ we have $$\sin7.5^\circ=\pm\sqrt{\frac{1-\cos15^\circ}{2}}$$ As $7.5^\circ$ is in quadrant I, $\sin7.5^\circ\gt0$, meaning we would select the positive square root. $$\sin7.5^\circ=\sqrt{\frac{1-\cos15^\circ}{2}}$$ *Now we need to calculate $\cos15^\circ$ As $15^\circ$ is half the angle $30^\circ$, $$\cos15^\circ=\cos\Big(\frac{30^\circ}{2}\Big)$$ Apply the following half-angle identity for cosine for $A=30^\circ$: $$\cos\Big(\frac{A}{2}\Big)=\pm\sqrt{\frac{1+\cos A}{2}}$$ we have $$\cos15^\circ=\pm\sqrt{\frac{1+\cos30^\circ}{2}}$$ As $15^\circ$ is in quadrant I, $\cos15^\circ\gt0$, meaning we would select the positive square root. $$\cos15^\circ=\sqrt{\frac{1+\cos30^\circ}{2}}$$ $$\cos15^\circ=\sqrt{\frac{1+\frac{\sqrt3}{2}}{2}}$$ $$\cos15^\circ=\sqrt{\frac{\frac{2+\sqrt3}{2}}{2}}$$ $$\cos15^\circ=\sqrt{\frac{2+\sqrt3}{4}}$$ $$\cos15^\circ=\frac{\sqrt{2+\sqrt3}}{2}$$ *Apply back to $\sin7.5^\circ$ $$\sin7.5^\circ=\sqrt{\frac{1-\frac{\sqrt{2+\sqrt3}}{2}}{2}}$$ $$\sin7.5^\circ=\sqrt{\frac{\frac{2-\sqrt{2+\sqrt3}}{2}}{2}}$$ $$\sin7.5^\circ=\sqrt{\frac{2-\sqrt{2+\sqrt3}}{4}}$$ $$\sin7.5^\circ=\frac{\sqrt{2-\sqrt{2+\sqrt3}}}{2}$$