Compute the infinite product [sin(x) cos(x/2)]^(1/2) * [sin(x/2) cos(x/4)]^(1/4) * [sin(x/4) cos(x/8)]^(1/8) * ... , where 0 <= x <= 2*Pi.

please tell me method too

Question

Compute the infinite product [sin(x) cos(x/2)]^(1/2) * [sin(x/2) cos(x/4)]^(1/4) * [sin(x/4) cos(x/8)]^(1/8) * ... , where 0 <= x <= 2*Pi. please tell me method too

Saurabh Koranglekar · Answer

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Compute the infinite product [sin(x) cos(x/2)]^(1/2) * [sin(x/2) cos(x/4)]^(1/4) * [sin(x/4) cos(x/8)]^(1/8) * ... , where 0 please tell me method too

Compute the infinite product [sin(x) cos(x/2)]^(1/2) * [sin(x/2) cos(x/4)]^(1/4) * [sin(x/4) cos(x/8)]^(1/8) * ... , where 0 <= x <= 2*Pi.

please tell me method too

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Compute the infinite product [sin(x) cos(x/2)]^(1/2) * [sin(x/2) cos(x/4)]^(1/4) * [sin(x/4) cos(x/8)]^(1/8) * ... , where 0 please tell me method too

Compute the infinite product [sin(x) cos(x/2)]^(1/2) * [sin(x/2) cos(x/4)]^(1/4) * [sin(x/4) cos(x/8)]^(1/8) * ... , where 0 <= x <= 2*Pi. please tell me method too

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Compute the infinite product [sin(x) cos(x/2)]^(1/2) * [sin(x/2) cos(x/4)]^(1/4) * [sin(x/4) cos(x/8)]^(1/8) * ... , where 0 <= x <= 2*Pi.

please tell me method too