Answer
$$\sin7.5^\circ=\frac{\sqrt{2-\sqrt{2+\sqrt3}}}{2}$$
Work Step by Step
$$\sin7.5^\circ$$
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 $\sin7.5^\circ$
$$\sin7.5^\circ=\sqrt{\frac{1-\frac{\sqrt{2+\sqrt3}}{2}}{2}}$$
$$\sin7.5^\circ=\sqrt{\frac{\frac{2-\sqrt{2+\sqrt3}}{2}}{2}}$$
$$\sin7.5^\circ=\sqrt{\frac{2-\sqrt{2+\sqrt3}}{4}}$$
$$\sin7.5^\circ=\frac{\sqrt{2-\sqrt{2+\sqrt3}}}{2}$$
