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Q.The sum ∑k=1nsin⁡kθ\sum_{k=1}^{n} \sin k\theta equals

a
sin⁡(nθ2)sin⁡((n+1)θ2)sin⁡(θ/2)\dfrac{\sin(\frac{n\theta}{2}) \sin(\frac{(n+1)\theta}{2})}{\sin(\theta/2)}
b
sin⁡(nθ2)cos⁡((n+1)θ2)sin⁡(θ/2)\dfrac{\sin(\frac{n\theta}{2}) \cos(\frac{(n+1)\theta}{2})}{\sin(\theta/2)}
c
sin⁡(nθ2)sin⁡(nθ2)sin⁡(θ/2)\dfrac{\sin(\frac{n\theta}{2}) \sin(\frac{n\theta}{2})}{\sin(\theta/2)}
d
cos⁡(θ2)sin⁡(nθ2)sin⁡(θ/2)\dfrac{\cos(\frac{\theta}{2}) \sin(\frac{n\theta}{2})}{\sin(\theta/2)}

Correct Answer: Option A

The correct solution involves applying the fundamental concept to derive the final value step by step...

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