1.

Prove that `sintheta+sin3theta+sin5theta+.....+sin(2n-1)theta=(sin^2ntheta)/(sintheta)dot`

Answer» Here, we will use,
`sinalpha+sin(alpha+beta)+sin(alpha+2beta)+...+sin(alpha+(n-1)beta) = (sin((nbeta)/2))/(sin(beta/2))**sin(alpha+((n-1)beta)/2)`
Now,
`L.H.S. = sintheta+sin3theta+sin5theta+...+sin(2n-1)theta`
`=sintheta+sin(theta+2theta)+sin(theta+2*2theta)+...sin(theta+(n-1)2theta)`
`=(sin((2ntheta)/2))/(sin((2theta)/2))**sin(theta+((n-1)2theta)/2)`
`=(sin(ntheta)/(sintheta))*sin(theta+ntheta-theta)`
`=sin^2(ntheta)/sintheta = R.H.S.`


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