The value of `sin( pi/n) + sin ((3pi)/n)+ sin ((5pi)/n)+...` to n terms is equal to

Answer

0

Explanation

We know that
`sin alpha+sin(alpha+beta)+...+sin{alpha+(n-1)beta}`
`=(sin{alpha+(n-1)(beta)/(2)}sin((n beta)/(2)))/(sin""(beta)/(2))`
`thereforesin""(pi)/(n)+sin""(3pi)/(n)+sin""(5pi)/(n)+....` to n terms
`=(sin{(pi)/(n)+(n-1)(pi)/(n)}sin((npi)/(n)))/(sin((pi)/(n)))" "[Here, alpha =(pi)/(n)and beta=(2pi)/(n)]`
`=(sinpui sin pi)/(sin""(pi)/(n))=0`