Given $$\cos {\pi \over {{2^2}}}.\cos {\pi \over {{2^3}}}\,.....\cos {\pi \over {{2^{10}}}}.\sin {\pi \over {{2^{10}}}}$$

Let $${\pi \over {{2^{10}}}}\, = \,\theta $$

$$ \therefore $$ $${\pi \over {{2^9}}}\, = \,2\theta $$

$${\pi \over {{2^8}}}\, = \,{2^2}\theta $$

$${\pi \over {{2^7}}}\, = \,{2^3}\theta $$
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$${\pi \over {{2^2}}}\, = \,{2^8}\theta $$

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 :

$$(\cos \theta .\cos 2\theta ......\cos {2^{n - 1}}\theta )$$ = $${{\sin {2^n}\theta } \over {{2^n}\sin \theta }}$$