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Question
Prove that, sinθ + sin2θ + sin3θ + ... + sinnθ = `((sin ntheta)/2 sin ((n + 1))/2 theta)/(sin theta/2)`, for all n ∈ N.
Solution
Let P(n): sinθ + sin2θ + sin3θ + ... + sinnθ
= `(sin (ntheta)/2 . sin ((n + 1))/2 theta)/(sin theta/2)`, n ∈ N.
Step 1: P(1) : sinθ = `(sin theta/2 . sin ((1 + 1)/2)theta)/(sin theta/2)`
= `(sin theta/2 . sin theta)/(sin theta/2)`
= sinθ
∴ sinθ = sinθ which is true for P(1).
Step 2: P(k): sinθ + sin2θ + sin3θ + ... + sinkθ
= `(sin (ktheta)/2 . sin ((k + 1)/2)theta)/(sin theta/2)`
Let it be true for P(k).
Step 3: P(k + 1): sinθ + sin2θ + sin3θ + ... + sin(k + 1)θ
= `(sin (ktheta)/2 . sin ((k + 1)/2)theta)/(sin theta/2) + sin(k + 1)theta`
= `(sin (ktheta)/2 . sin ((k + 1)/2)theta + sin(k + 1)theta . sin theta/2)/(sin theta/2)`
= `(2sin (ktheta)/2 . sin ((k + 1)/2)theta + 2 sin (k + 1)theta . sin theta/2)/(2 sin theta/2)`
= `(cos((ktheta)/2 - (k + 1)/2 theta) - cos((ktheta)/2 + (k + 1)/2 theta) + cos[(k + 1)theta - theta/2] - [cos[(k + 1)theta + theta/2]))/(2sin theta/2)`
= `(cos(- theta/2) - cos(ktheta + theta/2) + cos(ktheta + theta/2) - cos(ktheta + (3theta)/2))/(2sin theta/2)`
= `(cos(theta/2) - cos(ktheta + (3theta)/2))/(2sin theta/2)`
= `(-2sin((theta/2 + ktheta + (3theta)/2)/2).sin ((theta/2 - ktheta - (3theta)/2)/2))/(2sin theta/2)` ......`[because cos"A" - cos"B" = - 2sin (("A" + "B"))/2 sin (("A" - "B"))/2]`
= `(-2sin((ktheta + 2theta)/2) . sin ((-ktheta - theta)/2))/(2sin theta/2)`
= `(sin((ktheta + 2theta)/2).sin ((ktheta + theta)/2))/(sin theta/2)`
= `(sin[((k + 1) - 1)/2]theta.sin [(k + 1)/2]theta)/(sin theta/2)` which is true for P(k + 1).
Hence, P(k + 1) is true whenever P(k) is true.
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