Question

Real part of $\frac{1}{1 - cos\,\theta + i\,sin\,\theta}$ is

• A. $-\frac{1}{2}$
• B. $\frac{1}{2}$
• C. $\frac{1}{2} tan\,\theta/2$
• D. $2$

Answer 1

Answer: (B)

$z = \frac{1}{1 - cos\,\theta + i\,sin\,\theta}$
$ = \frac{1}{2\,sin^2\,\theta/2 + 2i\,sin\,\theta/2\,cos\,\theta/2}$
$ =\frac{1}{2i\,sin\,\theta/2} \frac{1}{(cos\,\theta/2 - i\,sin\,\theta/2)}$
$ = \frac{cos\,\theta/2 + i\,sin\,\theta/2}{2i\,sin\,\theta/2} = \frac{1}{2} + \frac{1}{2i} cot\,\theta/2$
$ = \frac{1}{2} - i \cdot \frac{1}{2} cot\,\theta/2$
Short Cut Method :
$z = \frac{1}{1 - cos\,\theta + i\,sin\,\theta} = \frac{1 - cos\,\theta - i\,sin\,\theta}{2( 1 -cos\,\theta)}$
$ = \frac{1 - cos\,\theta}{2( 1 - cos\,\theta)} - i \frac{sin\,\theta}{2(1 - cos\,\theta)}$
$\therefore $ Real part $ = 1/2$