Question

If the real part of the complex number ${(1-\cos \theta +2i\sin \theta )}^{-1}$ is $\frac{1}{5}$ for $\theta \in (0,\pi ),$ then the value of the integral $\underset{0}{\overset{\theta }{\int }}\sin xdx$ is equal to :

• A. $1$
• B. $2$
• C. $0$
• D. $-1$

Answer 1

The correct option is A $1$

$Z=\frac{1}{1-\cos \theta +2i\sin \theta }=\frac{(1-\cos \theta )-2i\sin \theta }{{(1-\cos \theta )}^2+4{\sin }^2\theta }$
$\therefore \text{Re}\left(Z\right)=\frac{1-\cos \theta }{2-2\cos \theta +3{\sin }^2\theta }=\frac{1}{5}$
$\therefore 5-5\cos \theta =2-2\cos \theta +3{\sin }^2\theta$
$\Rightarrow 3\cos \theta (1-\cos \theta )=0$
$\therefore \theta =\frac{\pi }{2},\text{when}\theta \in (0,\pi )$
$\therefore \underset{0}{\overset{\theta }{\int }}\sin xdx={\int }_0^{\frac{\pi }{2}}\sin xdx=1$