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Find sum of sin2alpha+sin3alpha+.......+...

Find sum of `sin2alpha+sin3alpha+.......+sinnalpha`, where `(n+2)alpha = 2 pi`

Text Solution

Verified by Experts

The correct Answer is:
`0`
`sum_(k=0)^(n-1)sin((2kpi)/h)=0`
`sin0+sin((2x)/h)+sin((4pi)/h)+...+sin((2(n-1))/npi)=0`
`alpha=(2pi)/(n+2)`
`sin2*(2pi)/(n+2)+sin3*(2pi)/(n+2)+...+sin n*(2pi)/(n+2)`
`sum_(k=0)^(n+1)sin((2kpi)/(n+2))=0`
`sin0+sin((2pi)/(n+2))+sin((4pi)/(n+2))+sin((6pi)/(n+2))+...+sin((2npi)/(n+2))+sin((2(n+2)pi)/(n+2))=0`
`f+sin(2pi)/(n+2)+sin((2(n+1)pi)/2)=0`
`sin((2(n+2-1)pi)/(n+2))`
...
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