# de moivre theorem and summation

• Apr 24th 2013, 02:11 AM
mathfag188
de moivre theorem and summation
By considering $\sum(1+\iota \tan \theta)^k$ from k=0 to k=n-1, show that
$\sum \cos k\theta \sec^k \theta = \cot \theta\sin n\theta\sec^n\theta$

stuck here

and i also dont know how to put limits in summation :P please tell that too
• Apr 24th 2013, 03:25 AM
BobP
Re: de moivre theorem and summation
Start by writing $(1+i\tan \theta)^{k}$ as

$\left(1+i \frac{\sin \theta}{\cos \theta}\right)^{k}=\frac{(\cos \theta + i \sin \theta)^{k}}{\cos^{k}\theta},$

and then make use of De Moivre's theorem.
• Apr 24th 2013, 07:01 AM
mathfag188
Re: de moivre theorem and summation
i already did what you told but cant go any further
can you please continue after that?
• Apr 24th 2013, 08:19 AM
BobP
Re: de moivre theorem and summation
If you do as suggested the summation can be split into two parts, one real the other imaginary.
The summation itself can be summed easily since it is a GP with common ratio $(1+i\tan \theta).$
Do that and then equate reals and imaginaries across the two results.
(Actually, you only need to equate reals.)