Dear Youssri,

I think that one can substitute $\theta=\pi/2+x$ and apply the well known inequality $|\sin nx|\leq n|\sin x|. 

Best regards,

Adolf Mirotin.

Dear Prof. Mirotin,

Thank you for your help.

Best wishes,

Youssri

Your inequality |cos(nx)| \leq |n*cos(x)| is true for all x and odd n. WMA n is positive. There is a simple inductive proof. Clearly true for n=1. Expand cos((n-2)+2)x) = cos((n-2)x)cos(2x)-sin((n-2)x)sin(2x). Use the triangle inequality and |sin(2x)|\leq 2|cos(x)|. Note that by induction, we have |cos((n-2)x)| \leq (n-2)|cos(x)|,  I prefer a direct proof to using second inequality that is just as easy/hard to verify.

Sincerely, Stephen

I appreciate your answer to my question. Thank you, Dr. Stephen.