1

Evaluate $$\sum_{k=1}^n\frac{\tan\frac x{2^k}}{2^{k-1} \cos\frac x{2^{k-1}}}$$ I am struck with calculation for this question

jimjim
  • 9,675
Samar Imam Zaidi
  • 8,800

1 Answers1

6

Hint:$$2\cot (2x)=\cot x-\tan x\\ \to \tan x=2\cot (2x)-\cot x$$so $$\frac{\tan\frac x{2^k}}{2^{k-1} \cos\frac x{2^{k-1}}}=\\ \frac{2\cot (2\frac x{2^k})-\cot \frac x{2^k}}{2^{k-1} \cos\frac x{2^{k-1}}}$$

Khosrotash
  • 24,922
  • @Robert Z I am trying to solve this question using above Hint but not able to proceed, i need your help – Samar Imam Zaidi Oct 21 '17 at 08:35
  • https://math.stackexchange.com/questions/3035402/calculation-of-complex-trigonometric-summation/3037970#3037970 – DXT Dec 16 '18 at 07:36