PMC Forum: value of n.

Question.....

Re: value of n.

by aditya ravichanndran - Friday, 22 July 2011, 5:47 PM

Firstly , we consider this equation

proof is easy ,we can use demoivre theorem and polynomials to prove it

if we put n=180 here

we get

$\sin(1).\sin(2)..\sin(89).\sin(90).\sin(91)...\sin(179)=\frac{180}{2^{179}} \\ \texttt{but} \ \sin(179)=\sin(1)..\sin(178)=\sin(2)...\texttt{so on}\\ \texttt{hence} \\ \left (\sin(1).\sin(2)..\sin(89) \right )^2= \frac{180}{2^{179}} \\ \boxed{\sin(1).\sin(2)..\sin(89)}=\sqrt{\frac{180}{2^{179}}}$

Re: value of n.

by jagdish singh - Sunday, 24 July 2011, 11:47 PM

thanks aditya ravichanndran

Physics, Chemistry, Mathematics Forums

Physics, Chemistry, Maths Forum

value of n.

value of n.

Re: value of n.

Re: value of n.