QJEEM19Ma8c8c: JEE Main 2019 Mathematics - Class 11 JEEM19Ma8c8c Solution

Question

Verified Solution • 2025–26 Session moderate

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 }}

Step-by-Step Verified Solution