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The sum S= sin theta + sin 2 theta + …+ ...

The sum `S= sin theta + sin 2 theta + …+ sin n theta, ` equals

A

`(sin""(ntheta)/(2)sin ""(theta(n+1))/(2))/(sin ""(theta)/(2))`

B

`(sin""(ntheta)/(2)cos ""(theta(n+1))/(2))/(sin ""(theta)/(2))`

C

`(cos""(ntheta)/(2)sin ""(theta(n+1))/(2))/(sin ""(theta)/(2))`

D

`(cos""(ntheta)/(2)cos ""(theta(n+1))/(2))/(sin ""(theta)/(2))`

Text Solution

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The correct Answer is:
A
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