1
$\begingroup$

On this thread, Bernard kindly gave me the formula

$$\sum_{k=0}^R \cos k \theta = \frac{\sin \frac{(R+1)\theta}{2}}{\sin \frac{\theta}{2}} \cos \frac{R \theta}{2}$$

He describes the formula as well-known. Does it have a name, so I can find a few pages relating to it?

Improve this question
$\endgroup$
  • $\begingroup$ It is an easy corollary of the Euler's formula. $\endgroup$ – Hanul Jeon Apr 21 '18 at 9:14
  • 1
    $\begingroup$ There is no $k$ in your summation. Anyway add the series with $i\sin(k\theta)$ and then it is a geometric series $(e^{i\theta})^k$ $\endgroup$ – zwim Apr 21 '18 at 9:15
  • $\begingroup$ I know no name. I simply learnt it when I was in last grade of high school. $\endgroup$ – Bernard Apr 21 '18 at 9:18
  • $\begingroup$ math.stackexchange.com/questions/17966/… $\endgroup$ – lab bhattacharjee Apr 21 '18 at 9:44
1
$\begingroup$

Lagrange Trigonometric Identity. See here

Improve this answer
$\endgroup$
  • $\begingroup$ Thank you. That link says Lagrange's identity is only valid for $0 \lt \theta \lt 2 \pi$. Is this true? I'm after a universal statement - or at least true for all real numbers. $\endgroup$ – Richard Burke-Ward Apr 21 '18 at 9:34
  • $\begingroup$ it is true. If $\theta \notin [0,2\pi ]$ that is not a problem since $\cos \left( k\theta \right)=\cos \left( k\theta +2n\pi \right)$ and you can choose $n$ such that $k\theta +2n\pi \in [0,2\pi ]$. $\endgroup$ – user547564 Apr 21 '18 at 11:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.