Solveeit Logo
Login

Question

Question: The real part of \((1 - \cos\theta + 2i\sin\theta)^{- 1}\) is...

The real part of (1−cos⁡θ+2isin⁡θ)−1(1 - \cos\theta + 2i\sin\theta)^{- 1} is

A

13+5cos⁡θ\frac{1}{3 + 5\cos\theta}

B

15−3cos⁡θ\frac{1}{5 - 3\cos\theta}

C

13−5cos⁡θ\frac{1}{3 - 5\cos\theta}

D

15+3cos⁡θ\frac{1}{5 + 3\cos\theta}

Answer

13−5cos⁡θ\frac{1}{3 - 5\cos\theta}

Explanation

Solution

Sol.{(1−cos⁡θ)+i.2sin⁡θ}−1={2sin⁡2θ2+i.4sin⁡θ2cos⁡θ2}−1\left\{ (1 - \cos\theta) + i.2\sin\theta \right\}^{- 1} = \left\{ 2\sin^{2}\frac{\theta}{2} + i.4\sin\frac{\theta}{2}\cos\frac{\theta}{2} \right\}^{- 1}

= (2sin⁡θ2)−1{sin⁡θ2+i.2cos⁡θ2}−1\left( 2\sin\frac{\theta}{2} \right)^{- 1}\left\{ \sin\frac{\theta}{2} + i.2\cos\frac{\theta}{2} \right\}^{- 1}

=(2sin⁡θ2)−1.1sin⁡θ2+i.2cos⁡θ2×sin⁡θ2−i.2cos⁡θ2sin⁡θ2−i.2cos⁡θ2=sin⁡θ2−i.2cos⁡θ22sin⁡θ2(sin⁡2θ2+4cos⁡2θ2)= \left( 2\sin\frac{\theta}{2} \right)^{- 1}.\frac{1}{\sin\frac{\theta}{2} + i.2\cos\frac{\theta}{2}} \times \frac{\sin\frac{\theta}{2} - i.2\cos\frac{\theta}{2}}{\sin\frac{\theta}{2} - i.2\cos\frac{\theta}{2}} = \frac{\sin\frac{\theta}{2} - i.2\cos\frac{\theta}{2}}{2\sin\frac{\theta}{2}\left( \sin^{2}\frac{\theta}{2} + 4\cos^{2}{}\frac{\theta}{2} \right)}

Hence, real part

=sin⁡θ22sin⁡θ2(1+3cos⁡2θ2)=12(1+3cos⁡2θ2)=15+3cos⁡θ= \frac{\sin\frac{\theta}{2}}{2\sin\frac{\theta}{2}\left( 1 + 3\cos^{2}\frac{\theta}{2} \right)} = \frac{1}{2\left( 1 + 3\cos^{2}\frac{\theta}{2} \right)} = \frac{1}{5 + 3\cos\theta}.