If e^{i \theta}=cis \theta then \sum_{n=0}^{\infty} \frac{\cos (n \theta)}{2^{n}}=

Correct Answer is : (4−2cosθ)(5−4cosθ)

(4+2 \cos \theta) /(5-4 \cos \theta)

(4-2 \cos \theta) /(5+4 \cos \theta)

(4-2 \cos \theta) /(5-4 \cos \theta)

(4+2 \cos \theta) /(5+4 \cos \theta)

AP EAPCET 04-Jul-2022 Shift 2 Paper

Section: Mathematics

Question : 10 of 160

Marks: +1, -0

If eiθ=cisθ then

∞

∑

n=0

‌

cos(nθ)

Go to Question:

Prev Question

Next Question