Question

$si{n}^2{24}^{\circ }-si{n}^2{6}^{\circ }=\frac{\sqrt{5}-1}{8}$

Answer 1

$L.H.S.si{n}^2{24}^{\circ }-si{n}^2{6}^{\circ }$
= sin (24 + 6) sin (24 - 6)

$[\because sin(A+B\left)sin\right(A-B)=si{n}^2-si{n}^{2}B]=sin{30}^{\circ }sin{18}^{\circ }=\frac{1}{2}.\frac{\sqrt{5}-1}{4}[\because sin{18}^{\circ }=\frac{\sqrt{5}-1}{4}]=\frac{\sqrt{5}-1}{8}$

= RHS