Question

How is sin2 +sin4+sin6+....sin88=
(Sin44.sin45)/sin1.....steps please. I am stuck.

Answer 1

If all angles are measured in degrees...\[S =\sum_{n=1}^{44}\sin(2n) \]\[2\sin (1)\cdot S = \sum_{n=1}^{44} 2\sin(1)\sin(2n)\]Now we use \(2 \sin A \sin B = \cos(A-B) - \cos(A+B) \) and see how quickly this turns into an easily manageable telescoping series.\[2\sin(1) \cdot S = 2\sin(1) \sin(2) + 2\sin(1)\sin(4) + \cdots + 2 \sin(1) \sin(88)\]\[=\cos(1) - \cos(3) + \cos (3) - \cos (5) + \cdots +\cos(87) - \cos(89)\]\[= \cos(1) - \cos(89)\]\[=2\sin(44) \sin(45) \]Therefore,\[\boxed{S= \frac{\sin(44) \sin(45)}{\sin(1)}}\]

Answer 2

Thnx