Given $\cos {\pi \over {{2^2}}}.\cos {\pi \over {{2^3}}}\,.....\cos {\pi \over {{2^{10}}}}.\sin {\pi \over {{2^{10}}}}$ Let ${\pi \over {{2^{10}}}}\, = \,\theta$ $\therefore$ ${\pi \over {{2^9}}}\, = \,2\theta$ ${\pi \over {{2^8}}}\, = \,{2^2}\theta$ ${\pi \over {{2^7}}}\, = \,{2^3}\theta$ . . ${\pi \over {{2^2}}}\, = \,{2^8}\theta$ So given term becomes, \cos {2^8}\theta .\cos {2^7}\theta .....\cos \theta $$$$.\sin {\pi \over {{2^{10}}}} = $(\cos \theta .\cos 2\theta ......\cos {2^8}\theta )\sin {\pi \over {{2^{10}}}}$ = ${{\sin {2^9}\theta } \over {{2^9}\sin \theta }}.\sin {\pi \over {{2^{10}}}}$ = ${{\sin {2^9}\left( {{\pi \over {{2^{10}}}}} \right)} \over {{2^9}\sin {\pi \over {{2^{10}}}}}}.\sin {\pi \over {{2^{10}}}}$ = ${{\sin \left( {{\pi \over 2}} \right)} \over {{2^9}}}$ = ${1 \over {{2^9}}}$ = ${1 \over {512}}$ Note : $(\cos \theta .\cos 2\theta ......\cos {2^{n - 1}}\theta )$ = ${{\sin {2^n}\theta } \over {{2^n}\sin \theta }}$