prove that $\sum_{k=1}^{n}{\sin(kA)}$

Question

prove that $$\sum_{k=1}^{n}{\sin(kA)} = {{\cos({A\over2})-\cos(nA+{A\over 2})}\over 2\sin({A\over 2})}$$using Telescoping series. How do i go about doing this?

Answer 1

Using the trigonometric identity: $$2\sin x \sin y = \cos(x - y) - \cos(x + y).$$ Multiply both sides by $2\sin(A/2)$, it follows that \begin{align*} & 2\sin(A/2)\sum_{k = 1}^{n} \sin(kA) \\ = & 2\sin(A/2)\sin(A) + 2\sin(A/2)\sin(2A) + \cdots + 2\sin(A/2)\sin(nA) \\ = & \cos(A/2 - A) - \cos(A/2 + A) + \cos(A/2 - 2A) + \cos(A/2 + 2A) + \cdots \\ & + \cos(A/2 - nA) + \cos(A/2 + nA) \\ = & \cos(A/2 - A) + \cos(A/2 + nA) \\ = & \cos(A/2) + \cos(A/2 + nA) \end{align*} where we used $\cos(x) = \cos(-x)$ to make the telescoping possible.