Question

The real part of $(1-\cos \left(\mathrm{\theta }\right)+2i\sin {\left(\mathrm{\theta }\right)}^{-1}$ is

• A. $\frac{1}{(3+5\cos \left(\theta \right))}$
• B. $\frac{1}{(5-3\cos \left(\theta \right))}$
• C. $\frac{1}{(3-5\cos \left(\theta \right))}$
• D. $\frac{1}{(5+3\cos \left(\theta \right))}$

Answer 1

The correct option is D

$\frac{1}{(5+3\cos \left(\theta \right))}$

Compute the required value.

Step 1: Simplify the expression

Given : ${\left[1-\cos \left(\mathrm{\theta }\right)+2i\sin \left(\mathrm{\theta }\right)\right]}^{-1}$

$$\begin{gathered} \Rightarrow {\left[2{\sin }^2\left(\frac{\mathrm{\theta }}{2}\right)+2\mathrm{i}\sin \left(\mathrm{\theta }\right)\right]}^{-1} \\ \Rightarrow {\left[2{\sin }^2\left(\frac{\mathrm{\theta }}{2}\right)+2\mathrm{i}\sin \left(\frac{\mathrm{\theta }}{2}\right)\cos \left(\frac{\mathrm{\theta }}{2}\right)\right]}^{-1} \\ \Rightarrow {\left[2\sin \left(\frac{\mathrm{\theta }}{2}\right)\right]}^{-1}{\left[\sin \left(\frac{\mathrm{\theta }}{2}\right)+\mathrm{i}2\cos \left(\frac{\mathrm{\theta }}{2}\right)\right]}^{-1} \\ \Rightarrow \frac{\left[\sin \left(\frac{\mathrm{\theta }}{2}\right)-\mathrm{i}2\cos \left(\frac{\mathrm{\theta }}{2}\right)\right]}{\left[2\sin \left(\frac{\mathrm{\theta }}{2}\right)\right]\left[\sin \left(\frac{\mathrm{\theta }}{2}\right)+\mathrm{i}2\cos \left(\frac{\mathrm{\theta }}{2}\right)\right]\left[\sin \left(\frac{\mathrm{\theta }}{2}\right)-\mathrm{i}2\cos \left(\frac{\mathrm{\theta }}{2}\right)\right]} \\ \Rightarrow \frac{\left[\sin \left(\frac{\mathrm{\theta }}{2}\right)-\mathrm{i}2\cos \left(\frac{\mathrm{\theta }}{2}\right)\right]}{\left[2\sin \left(\frac{\mathrm{\theta }}{2}\right)\right]\left[{\sin }^2\left(\frac{\mathrm{\theta }}{2}\right)+4{\cos }^2\left(\frac{\mathrm{\theta }}{2}\right)\right]} \\ \Rightarrow \frac{\left[\sin \left(\frac{\mathrm{\theta }}{2}\right)-\mathrm{i}2\cos \left(\frac{\mathrm{\theta }}{2}\right)\right]}{\left[2\sin \left(\frac{\mathrm{\theta }}{2}\right)\right]\left[1+3{\cos }^2\left({\displaystyle \frac{\mathrm{\theta }}{2}}\right)\right]} \\ \Rightarrow \frac{\left[\sin \left(\frac{\mathrm{\theta }}{2}\right)\right]}{\left[2\sin \left(\frac{\mathrm{\theta }}{2}\right)\right]\left[1+3{\cos }^2\left({\displaystyle \frac{\mathrm{\theta }}{2}}\right)\right]}-\frac{\mathrm{i}\left[2\cos \left(\frac{\mathrm{\theta }}{2}\right)\right]}{\left[2\sin \left(\frac{\mathrm{\theta }}{2}\right)\right]\left[1+3{\cos }^2\left({\displaystyle \frac{\mathrm{\theta }}{2}}\right)\right]} \end{gathered}$$

Step 2: Separate the real part

So real part is,

$$\begin{gathered} \frac{\left[\sin \left(\frac{\mathrm{\theta }}{2}\right)\right]}{\left[2\sin \left(\frac{\mathrm{\theta }}{2}\right)\right]\left[1+3{\cos }^2\left({\displaystyle \frac{\mathrm{\theta }}{2}}\right)\right]} \\ \Rightarrow \frac{1}{2}\times \frac{1}{\left[1+3{\displaystyle \frac{\left(1+\cos \left(\theta \right)\right)}{2}}\right]} \\ \Rightarrow \frac{1}{2}\times \frac{1}{\left[{\displaystyle \frac{2+3+3\cos \left(\theta \right)}{2}}\right]} \\ \Rightarrow \frac{1}{\left[5+3\cos \left(\theta \right)\right]} \end{gathered}$$

Therefore, real part is $\frac{1}{\left[5+3\cos \left(\theta \right)\right]}$.

Hence option (D) is the correct option.