Summation Of Series In Trigonometry - Practice Questions & MCQ

Trigonometric Series

$\begin{array}{c}\sin (\alpha )+\sin (\alpha +\beta )+\sin (\alpha +2\beta )+\sin (\alpha +3\beta )+\dots \dots \dots \dots \dots +\sin (\alpha +(n-1)\beta )\\ =\frac{\sin \left(\frac{\mathrm{n}\beta }{2}\right)\cdot \sin [\alpha +(\mathrm{n}-1)\frac{\beta }{2}]}{\sin \left(\frac{\beta }{2}\right)}\end{array}$
Proof:
Let $S=\sin \alpha +\sin (\alpha +\beta )+\sin (\alpha +2\beta )+\cdots +\sin (\alpha +\overline{n-1}\beta )$
Here angle are in A.P. and common difference of angles $=\beta$ multiplying both side by $2\sin \frac{\beta }{2}$ we get,

$\begin{array}{rl}& 2S\sin \frac{\beta }{2}=2\sin \alpha \sin \frac{\beta }{2}+2\sin (\alpha +\beta )\sin \frac{\beta }{2}+\cdots +2\sin (\alpha +\overline{n-1}\beta )\sin \frac{\beta }{2}\\ & \text{ Now, }2\sin \alpha \sin \frac{\beta }{2}=\cos (\alpha -\frac{\beta }{2})-\cos (\alpha +\frac{\beta }{2})\\ & 2\sin (\alpha +\beta )\sin \frac{\beta }{2}=\cos (\alpha +\frac{\beta }{2})-\cos (\alpha +\frac{3\beta }{2})\\ & 2\sin (\alpha +2\beta )\sin \frac{\beta }{2}=\cos (\alpha +\frac{3\beta }{2})-\cos (\alpha +\frac{\beta \beta }{2})\\ & \dots \\ & \cdots \\ & \cdots \\ & 2\sin (\alpha +\overline{n-1}\beta )\sin \frac{\beta }{2}=\cos [\alpha +(2n-3)\frac{\beta }{2}]-\cos [\alpha (2n-1)\frac{\beta }{2}]\end{array}$

Adding all the above we get

$\begin{array}{rl}& 2\sin \frac{\beta }{2}\mathrm{S}=\cos (\alpha -\frac{\beta }{2})-\cos (\alpha +(2\mathrm{n}-1)\frac{\beta }{2})\\ & \text{ or }\\ & 2\sin \frac{\beta }{2}\mathrm{S}=2\sin (\alpha +(\mathrm{n}-1)\frac{\beta }{2})\sin \frac{\mathrm{n}\beta }{2}\\ & \Rightarrow \mathrm{S}=\frac{\sin (\alpha +(\mathrm{n}-1)\frac{\beta }{2})\sin \frac{\mathrm{n}\beta }{2}}{\sin \frac{\beta }{2}}\end{array}$
For cosine series

$\cos \alpha +\cos (\alpha +\beta )+\cos (\alpha +2\beta )+\cdots +\cos (\alpha +\overline{n-1}\beta )=\frac{\sin \frac{n\beta }{2}}{\sin \frac{\beta }{2}}\cos [\alpha +(n-1)\frac{\beta }{2}]$