Question

Prove that:
${\sin }^2\frac{2\pi }{5}-{\sin }^{2-}\frac{\pi }{3}=\frac{\sqrt{5}{\displaystyle -}{\displaystyle 1}}{8}$

Answer 1

$$\begin{gathered} \frac{2\pi }{5}=72°,\frac{\pi }{3}=60° \\ \mathrm{LHS}={\sin }^{2}72°-{\sin }^{2}60° \\ ={\sin }^2\left(90°-18°\right)-\frac{3}{4} \\ ={\cos }^{2}18°-\frac{3}{4}\left(\because \sin \left(90°-\mathrm{\theta }\right)=\mathrm{cos\theta }\right) \\ ={\left(\frac{\sqrt{10+2\sqrt{5}}}{4}\right)}^2-\frac{3}{4}\left(\because \cos 18°=\frac{\sqrt{10+2\sqrt{5}}}{4}\right) \\ =\frac{10+2\sqrt{5}}{16}-\frac{3}{4} \end{gathered}$$

$$\begin{gathered} =\frac{10+2\sqrt{5}-12}{16} \\ =\frac{2\sqrt{5}-2}{16} \\ =\frac{\sqrt{5}-1}{8} \\ =\mathrm{RHS} \\ \mathrm{Hence}\mathrm{proved}. \end{gathered}$$