© All Rights reserved @ LearnWithDash
Step-by-Step Solution
Step 1: Express the given complex number
Let
$z = \frac{1}{1 - \cos\theta + 2i \sin\theta}.$
Our goal is to determine the condition on $\theta$ such that the real part of $z$ is $\tfrac{1}{5}$, where $\theta \in (0,\pi)$.
Step 2: Simplify the denominator
Use the trigonometric identities:
$1 - \cos\theta = 2 \sin^2\left(\tfrac{\theta}{2}\right)$
$\sin\theta = 2 \sin\left(\tfrac{\theta}{2}\right)\cos\left(\tfrac{\theta}{2}\right)$
Then,
$1 - \cos\theta + 2i \sin\theta
= 2 \sin^2\left(\frac{\theta}{2}\right)
+ 2i \cdot 2 \sin\left(\frac{\theta}{2}\right)\cos\left(\frac{\theta}{2}\right).$
Step 3: Find the real part of $z$
To find $\mathrm{Re}(z)$, one usually multiplies numerator and denominator by the complex conjugate of the denominator, or uses known factorizations. After simplification, it is found that
$\mathrm{Re}(z)
= \frac{1}{2\Big(\sin^2\left(\frac{\theta}{2}\right) + 4\cos^2\left(\frac{\theta}{2}\right)\Big)}.$
We are given $\mathrm{Re}(z) = \tfrac{1}{5}$. Hence,
$\frac{1}{2\Big(\sin^2\left(\frac{\theta}{2}\right) + 4\cos^2\left(\frac{\theta}{2}\right)\Big)} = \frac{1}{5}.$
This implies
$\sin^2\left(\frac{\theta}{2}\right) + 4\cos^2\left(\frac{\theta}{2}\right) = \frac{5}{2}.$
Step 4: Solve for $\theta$
Using
$\sin^2\left(\frac{\theta}{2}\right) = 1 - \cos^2\left(\frac{\theta}{2}\right),$
we rewrite:
$\bigl(1 - \cos^2\left(\frac{\theta}{2}\right)\bigr) + 4\cos^2\left(\frac{\theta}{2}\right)
= \frac{5}{2}.$
Hence,
$1 + 3\cos^2\left(\frac{\theta}{2}\right) = \frac{5}{2},$
$3\cos^2\left(\frac{\theta}{2}\right) = \frac{3}{2},$
$\cos^2\left(\frac{\theta}{2}\right) = \frac{1}{2}.$
Therefore,
$\cos\left(\frac{\theta}{2}\right) = \frac{1}{\sqrt{2}}
\quad \text{or} \quad
\cos\left(\frac{\theta}{2}\right) = -\frac{1}{\sqrt{2}}.$
Since $\theta \in (0,\pi)$, we have $\tfrac{\theta}{2} \in (0, \tfrac{\pi}{2})$. In this interval, $\cos\left(\frac{\theta}{2}\right)$ is positive, so
$\cos\left(\frac{\theta}{2}\right) = \frac{1}{\sqrt{2}} \implies \frac{\theta}{2} = \frac{\pi}{4} \implies \theta = \frac{\pi}{2}.$
Step 5: Evaluate the integral
We need to find
$\displaystyle \int_0^\theta \sin x\,dx$
and we have just determined $\theta = \frac{\pi}{2}.$ So,
$\int_0^{\pi/2} \sin x\,dx = \left[-\cos x\right]_0^{\pi/2} = -\cos\left(\frac{\pi}{2}\right) + \cos(0) = 0 + 1 = 1.$
Final Answer
$\displaystyle 1$
