Question

Prove that ${\sin }^2{72}^{\circ }-{\sin }^2{60}^{\circ }=\frac{\sqrt{5}-1}{8}$

Answer 1

$\sin {72}^0=\frac{\sqrt{10+2\sqrt{5}}}{4}$ and $\sin {60}^0=\frac{\sqrt{3}}{2}$

$⟹\frac{10+2\sqrt{5}}{16}-\frac{3}{4}$

$=\frac{2\sqrt{5}-2}{16}$

$=\frac{\sqrt{5}-1}{8}$

$\because$ it is proved