STD 11 - 3. trignometrical ratios,functions and identities — JEE STD 11 Science — Question

$ = \frac{1}{2}\sin 20^\circ \sin 60^\circ \,(2\sin {40^o}\sin 80^\circ )$

$ = \frac{1}{2}\sin 20^\circ \sin 60^\circ (\cos 40^\circ - \cos 120^\circ )$

$ = \frac{1}{2}.\frac{{\sqrt 3 }}{2}\sin 20^\circ \left( {1 - 2{{\sin }^2}20^\circ + \frac{1}{2}} \right)$

$ = \frac{{\sqrt 3 }}{4}\sin 20^\circ \left( {\frac{3}{2} - 2{{\sin }^2}20^\circ } \right)$

$ = \frac{{\sqrt 3 }}{8}(3\sin 20^\circ - 4{\sin ^3}20^\circ )$

$ = \frac{{\sqrt 3 }}{8}\sin 60^\circ = \frac{{\sqrt 3 }}{8}.\frac{{\sqrt 3 }}{2} = \frac{3}{{16}}$.