Real part of (1- cos θ+2 i sin θ)-1 is

Question

Real part of (1- cos θ+2 i sin θ)-1 is (A) (1/(3+5 cos θ)) (B) (1/(5+3 cos θ)) (C) (1/(5-3 cos θ)) (D) (1/(3-5 cos θ))

Tardigrade Admin · Accepted Answer

z =(1/1- cos θ+2 i sin θ) =(1- cos θ-2 i sin θ/(1- cos θ)2+4 sin 2 θ) Real part =(1- cos θ/(1- cos θ)2+4(1- cos 2 θ)) =(1/1- cos θ+4(1+ cos θ)) =(1/5+3 cos θ)

Q. Real part of $(1 - cos θ + 2 i sin θ)^{- 1}$ is

$\frac{1}{( 3 + 5 c o s θ )}$

$\frac{1}{( 5 + 3 c o s θ )}$

$\frac{1}{( 5 - 3 c o s θ )}$

$\frac{1}{( 3 - 5 c o s θ )}$

$z = \frac{1}{1 - c o s θ + 2 i s i n θ}$
$= \frac{1 - c o s θ - 2 i s i n θ}{( 1 - c o s θ ) ^{2} + 4 s i n ^{2} θ}$
Real part $= \frac{1 - c o s θ}{( 1 - c o s θ ) ^{2} + 4 ( 1 - c o s ^{2} θ )}$
$= \frac{1}{1 - c o s θ + 4 ( 1 + c o s θ )}$
$= \frac{1}{5 + 3 c o s θ}$

Q. Real part of (1−cosθ+2isinθ)−1 is

(3+5cosθ)1​

(5+3cosθ)1​

(5−3cosθ)1​

(3−5cosθ)1​