Question

Prove that:
${\sin }^{2}42°-{\cos }^{2}78=\frac{\sqrt{5}+1}{8}$

Answer 1

$$\begin{gathered} \mathrm{LHS}={\sin }^{2}42°-{\cos }^{2}78° \\ ={\sin }^2\left(90°-48°\right)-{\cos }^2\left(90°-12°\right) \\ ={\cos }^{2}48°-{\sin }^{2}12° \\ =\cos \left(48°+12°\right)\cos \left(48°-12°\right)\left[\cos \left(\mathrm{A}+\mathrm{B}\right)\cos \left(\mathrm{A}-\mathrm{B}\right)={\cos }^2\mathrm{A}-{\sin }^2\right] \\ =\cos 60°\cos 36° \\ =\frac{1}{2}\times \frac{\sqrt{5}+1}{4}\left(\because \cos 36°=\frac{\sqrt{5}+1}{4}\right) \\ =\frac{\sqrt{5}+1}{8} \\ =\mathrm{RHS} \\ \mathrm{Hence}\mathrm{proved}. \end{gathered}$$