example of telescoping sum

Some trigonometric sums, as ${\sum }_{k=1}^n\cos k\alpha$ and ${\sum }_{k=1}^n\sin k\alpha$, may be telescoped if the terms are first edited by a suitable goniometric formula (http://planetmath.org/GoniometricFormulae) (‘‘product formula’’). E.g. we may write:

$$\sum _{k=1}^n\cos k\alpha =\frac{1}{\sin \frac{\alpha }{2}}\sum _{k=1}^n\cos k\alpha \sin \frac{\alpha }{2}$$

The product formula  $\cos x\sin y=\frac{1}{2}[\sin (x+y)-\sin (x-y)]$  alters this to

$$\sum _{k=1}^n\cos k\alpha =\frac{1}{2\sin \frac{\alpha }{2}}\sum _{k=1}^n\left(\sin \frac{(2k+1)\alpha }{2}-\sin \frac{(2k-1)\alpha }{2}\right),$$

or

$$\sum _{k=1}^n\cos k\alpha =\frac{1}{2\sin \frac{\alpha }{2}}(\sin \frac{3\alpha }{2}-\sin \frac{\alpha }{2}+\sin \frac{5\alpha }{2}-\sin \frac{3\alpha }{2}+-\mathrm{\dots }+\sin \frac{(2n+1)\alpha }{2}-\sin \frac{(2n-1)\alpha }{2}).$$

After cancelling the opposite numbers we obtain the formula

${\displaystyle \sum _{k=1}^n}\cos k\alpha ={\displaystyle \frac{\sin \frac{(2n+1)\alpha }{2}-\sin \frac{\alpha }{2}}{2\sin \frac{\alpha }{2}}}.$(1)

The corresponding formula

${\displaystyle \sum _{k=1}^n}\sin k\alpha ={\displaystyle \frac{-\cos \frac{(2n+1)\alpha }{2}+\cos \frac{\alpha }{2}}{2\sin \frac{\alpha }{2}}}.$(2)

is derived analogously.

Note.  The formulae (1) and (2) are gotten also by adding the left side of the former and $i$ times the left side of the latter and then applying de Moivre identity.