Question

The real part of $\frac {1} {1+Cos \, \theta+i \, Sin \, \theta}$ is

• A. $-\frac {1}{2}$
• B. $\frac {1}{2}$
• C. $\sqrt {2}$
• D. $\frac {1} {\sqrt{2}}$

Answer 1

Answer: (B)

Let $z =\frac{1}{1+\cos \theta+i \sin \theta} $
$=\frac{1}{2 \cos ^{2} \frac{\theta}{2}+2 i \sin \frac{\theta}{2} \cos \frac{\theta}{2}} $
$=\frac{1}{2 \cos \frac{\theta}{2}\left(\cos \frac{\theta}{2}+i \sin \frac{\theta}{2}\right)} $
$=\frac{1}{2 \cos \frac{\theta}{2}}\left(\cos \frac{\theta}{2}+i \sin \frac{\theta}{2}\right)^{-1} $
$=\frac{\cos \left(-\frac{\theta}{2}\right)+i \sin \left(-\frac{\theta}{2}\right)}{2 \cos \frac{\theta}{2}}=\frac{\cos \frac{\theta}{2}-i \sin \frac{\theta}{2}}{2 \cos \frac{\theta}{2}}$
[using De-Moivre's Theorem]
$=\frac{1}{2}\left[1-i \tan \frac{\theta}{2}\right]$
Real of $z=\frac{1}{2}$ real part of $\left[1-i \tan \frac{\theta}{2}\right]$
$=\frac{1}{2}(1)=\frac{1}{2}$