The value of [ 2^2. 2^3,..... 2^10. 2^10] is -

Question

The value of $$\cos {\pi  \over {{2^2}}}.\cos {\pi  \over {{2^3}}}\,.....\cos {\pi  \over {{2^{10}}}}.\sin {\pi  \over {{2^{10}}}}$$ is -

Accepted Answer

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 }}$

Solutions · 1