Question

Prove that:
${\sin }^{2}24°-{\sin }^{2}6°=\frac{\sqrt{5}-1}{8}$

Answer 1

$$\begin{gathered} \mathrm{LHS}={\sin }^{2}24°-{\sin }^{2}6° \\ =\sin \left(24°+6°\right)\sin \left(24°-6°\right)\left[\sin \left(\mathrm{A}+\mathrm{B}\right)\sin \left(\mathrm{A}-\mathrm{B}\right)={\sin }^2\mathrm{A}-{\sin }^2\mathrm{B}\right] \\ =\sin 30°\sin 18° \\ =\frac{1}{2}\times \frac{\sqrt{5}-1}{4}\left(\because \sin 18°=\frac{\sqrt{5}-1}{4}\right) \\ =\frac{\sqrt{5}-1}{8} \\ =\mathrm{RHS} \\ \mathrm{Hence}\mathrm{proved}. \end{gathered}$$