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Exercise 4.4 · Q15

Q.Prove using the Mathematical induction
[!FORMULA] sin⁡(α)+sin⁡(α+π6)+sin⁡(α+2π6)+⋯+sin⁡(α+(n−1)π6)=sin⁡(α+(n−1)π12)×sin⁡(nπ12)sin⁡(π12).\sin(\alpha)+\sin\left(\alpha+\dfrac{\pi}{6}\right)+\sin\left(\alpha+\dfrac{2\pi}{6}\right)+\cdots+\sin\left(\alpha+\dfrac{(n-1)\pi}{6}\right) = \dfrac{\sin\left(\alpha+\dfrac{(n-1)\pi}{12}\right)\times\sin\left(\dfrac{n\pi}{12}\right)}{\sin\left(\dfrac{\pi}{12}\right)}.

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Let P(n):sin⁡α+sin⁡(α+π6)+⋯+sin⁡(α+(n−1)π6)=sin⁡(α+(n−1)π12)sin⁡(nπ12)sin⁡(π12)P(n):\sin\alpha+\sin(\alpha+\tfrac\pi6)+\cdots+\sin\big(\alpha+\tfrac{(n-1)\pi}6\big)=\dfrac{\sin\left(\alpha+\tfrac{(n-1)\pi}{12}\right)\sin\left(\tfrac{n\pi}{12}\right)}{\sin\left(\tfrac\pi{12}\right)}. This mirrors Example 4.69's cosine-series proof exactly, with β=π/6\beta=\pi/6.

Step 1. Base case. P(1)P(1): LHS =sin⁡α=\sin\alpha; RHS =sin⁡(α)sin⁡(π/12)sin⁡(π/12)=sin⁡α=\dfrac{\sin(\alpha)\sin(\pi/12)}{\sin(\pi/12)}=\sin\alpha. True.

Step 2. Inductive hypothesis. Assume P(k)P(k) holds with sum to the kkth term equal to sin⁡(α+(k−1)π12)sin⁡(kπ12)sin⁡(π12)\dfrac{\sin\left(\alpha+\tfrac{(k-1)\pi}{12}\right)\sin\left(\tfrac{k\pi}{12}\right)}{\sin\left(\tfrac\pi{12}\right)}. …

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