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

Question

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

Tardigrade Admin · Accepted Answer

z = (1/1 - cos θ + i sin θ) = (1/2 sin2 θ/2 + 2i sin θ/2 cos θ/2) =(1/2i sin θ/2) (1/(cos θ/2 - i sin θ/2)) = (cos θ/2 + i sin θ/2/2i sin θ/2) = (1/2) + (1/2i) cot θ/2 = (1/2) - i ⋅ (1/2) cot θ/2 <bR> Short Cut Method: z = (1/1 - cos θ + i sin θ) = (1 - cos θ - i sin θ/2( 1 -cos θ)) = (1 - cos θ/2( 1 - cos θ)) - i (sin θ/2(1 - cos θ)) ∴ Real part = 1/2

Q. Real part of $\frac{1}{1 - cos θ + i s in θ}$ is

$- \frac{1}{2}$

$\frac{1}{2}$

$\frac{1}{2} t an θ /2$

$2$

Q. Real part of 1−cosθ+isinθ1​ is

−21​

21​

21​tanθ/2

2

Q. Real part of $\frac{1}{1 - cos θ + i s in θ}$ is

$- \frac{1}{2}$

$\frac{1}{2}$

$\frac{1}{2} t an θ /2$

$2$