Half-Angle Values

Put x/2 in place of x in the power-reducing formulas and they find the values at half an angle you know:

$${\sin\left(\frac{x}{2}\right)}^{2} = \frac{1-\cos\left(x\right)}{2}$$

$${\cos\left(\frac{x}{2}\right)}^{2} = \frac{1+\cos\left(x\right)}{2}$$

15° is half of 30°. So ${\sin\left(15^{\circ}\right)}^{2} = \frac{1-\cos\left(30^{\circ}\right)}{2} = \frac{1-\frac{\sqrt{3}}{2}}{2}$. Take the half out of the top: $\frac{\frac{1}{2}\left(2-\sqrt{3}\right)}{2}$, which is $\frac{2-\sqrt{3}}{4}$.

The formula gives the square. To get sin 15° itself take the square root and choose the sign from the quadrant: 15° is in quadrant I, so the positive root, $\sin\left(15^{\circ}\right) = \frac{\sqrt{2-\sqrt{3}}}{2}$.