Using the formulae for \(\sin(A\pm B)\), prove that \(\sin(r+\frac12)\theta-\sin(r-\frac12)\theta\equiv2\cos r\theta\sin\frac12\theta\).[2]
Hence find a formula for \(\displaystyle\sum_{r=1}^n\cos r\theta\) in terms of \(\sin(n+\frac12)\theta\) and \(\sin\frac12\theta\).[3]
Prove by the method of mathematical induction that \(\displaystyle\sum_{r=1}^n\sin r\theta=\frac{\cos\frac12\theta-\cos(n+\frac12)\theta}{2\sin\frac12\theta}\) for all positive integers \(n\).[6]
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Answer:(i) Proven. (ii) \(\frac{\sin((n+1/2)\theta)-\sin(\theta/2)}{2\sin(\theta/2)}\). (iii) Proven by induction, where the quotient is defined.