Trigonometric Product

[stuart clark]

How can I calculate

[alpercay]

sin(pi/2n+1).sin(2pi/2n+1).sin(3pi/2n+1)...sin(npi/2n+1) = sqrtr(2n+1)/2ⁿ

To see proof one can look up "Ussr Olympiad Problem Book,problem 231,page 334.But an elegant solution may be exist via using trigonometric identities, complex numbers or geometry.It is possible that calculate value of sin(pi/7) on triangle with 3pi/7,3pi/7 and pi/7 angles, but rather dull.Answer sqrt(7)/8.

[matsever44]

fikir verebilir

[stuart clark]

Thanks Admin and Matsever44