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

$ = {\sin ^2}{36^o}\,\,{\sin ^2}\,{72^o} = \frac{1}{4}\,\left\{ {(2\,\,{{\sin }^2}{{36}^o})\,\,(2\,\,{{\sin }^2}\,\,{{72}^o})} \right\}$

$ = \frac{1}{4}\left\{ {(1 - \cos \,\,{{72}^o})\,\,(1 - \cos \,\,{{144}^o})} \right\}$

$ = \frac{1}{4}\left\{ {(1 - \sin \,\,{{18}^o})\,\,(1 + \cos \,\,{{36}^o})} \right\}$

$ = \frac{1}{4}\left[ {\left( {1 - \frac{{\sqrt 5 - 1}}{4}} \right)\,\,\left( {1 + \frac{{\sqrt 5 + 1}}{4}} \right)} \right]$

$= \frac{{20}}{{16}} \times \frac{1}{4} = \frac{5}{{16}}$.