Question

Given:-
cos(x/2)*cos(x/4)*cos(x/8)...=sinx/x
then, prove that:-
(1/2^2)sec^2(x/2)+(1/4^2)sec^2(x/4)+(1/8^2)sec^2(x/8)...=Cosec^2*1/x^2.​

Answer 1

Answer:

Given cos2xcos4xcos8x⋯=xsinx

∴log(cos2x)+log(cos4x)+⋯

=log(sinx)−logx 

Differentiating twice both sides we get 

221sec22x+241sec24x+⋯=cosec2x−x21