Question

If $\mathrm{Re}\left[{(1+\cos \theta +2i\sin \theta )}^{-1}\right]=4$, then the value of $\theta$ is

• A. $\frac{\pi }{2}$
• B. $\frac{\pi }{3}$
• C. $-\frac{\pi }{3}$
• D. $\pi$

Answer 1

The correct option is D

$\pi$

The explanation for the correct option:

Step 1: Simplification of the expression

Here, the expression can be rewritten as $\frac{1}{1+\cos \left(\theta \right)+2i\sin \left(\theta \right)}$.

Now, multiply the numerator and denominator of the above expression by the conjugate of its denominator,

$\begin{array}{c}\frac{1}{\left(1+\cos \left(\theta \right)+2i\sin \left(\theta \right)\right)}=\frac{1}{\left(1+\cos \left(\theta \right)+2i\sin \left(\theta \right)\right)}\times \frac{\left(1+\cos \left(\theta \right)-2i\sin \left(\theta \right)\right)}{\left(1+\cos \left(\theta \right)-2i\sin \left(\theta \right)\right)}\\ =\frac{\left(1+\cos \left(\theta \right)-2i\sin \left(\theta \right)\right)}{{\left(1+\cos \left(\theta \right)\right)}^2-{\left(2i\sin \left(\theta \right)\right)}^2}\\ =\frac{\left(1+\cos \left(\theta \right)-2i\sin \left(\theta \right)\right)}{1+{\cos }^2\theta +2\cos \theta +4{\sin }^2\theta }\left[<span\; style="font-family:"Cambria\; Math",serif;mso-bidi-font-family:"Cambria\; Math";">\because </span>{\left(a+b\right)}^2=a^2+b^2+2ab\right]\\ =\frac{\left(1+\cos \left(\theta \right)-2i\sin \left(\theta \right)\right)}{1+{\cos }^2\theta +2\cos \theta +4\left(1-{\cos }^2\theta \right)}\left[<span\; style="font-family:"Cambria\; Math",serif;mso-bidi-font-family:"Cambria\; Math";">\because </span>{\sin }^2\left(\theta \right)+{\cos }^2\left(\theta \right)=1\right]\\ =\frac{\left(1+\cos \left(\theta \right)-2i\sin \left(\theta \right)\right)}{1+{\cos }^2\theta +2\cos \theta +4-4{\cos }^2\theta }\\ =\frac{\left(1+\cos \left(\theta \right)-2i\sin \left(\theta \right)\right)}{5+2\cos \left(\theta \right)-3{\cos }^2\left(\theta \right)}\end{array}$

The real part of ${\left(1+\cos \left(\theta \right)+2i\sin \left(\theta \right)\right)}^{-1}$ is $\frac{1+\cos \left(\theta \right)}{5+2\cos \left(\theta \right)-3{\cos }^2\left(\theta \right)}$.

Step 2: Equate left-hand side with right-hand side, then solve for $\theta$

Since $\mathrm{Re}\left[{\left(1+\cos \left(\theta \right)+2i\sin \left(\theta \right)\right)}^{-1}\right]=4$, then we have

$\frac{1+\cos \left(\theta \right)}{5+2\cos \left(\theta \right)-3{\cos }^2\left(\theta \right)}=4$

Substitute $\cos \left(\theta \right)$ by $t$,

$\begin{array}{rcl}\therefore \frac{1+t}{5+2t-3t^2}& =& 4\\ & \Rightarrow & 1+t=20+8t-12t^2\\ & \Rightarrow & 12t^2-7t-19=0\end{array}$

Solve the equation by splitting the middle term,

$\begin{array}{rcl}\therefore 12t^2-7t-19& =& 0\\ & \Rightarrow & 12t^2-19t+12t-19=0\\ & \Rightarrow & t\left(12t-19\right)+1\left(12t-19\right)=0\\ & \Rightarrow & \left(t+1\right)\left(12t-19\right)=0\end{array}$

Replace $t$ by $\cos \left(\theta \right)$, we get

$\cos \left(\theta \right)+1=0\text{and}12\cos \left(\theta \right)-19=0$

From this, $12\cos \left(\theta \right)-19=0$ can be neglected as it is not required.

Then, we have

$\begin{array}{rcl}\cos \left(\theta \right)& =& -1\end{array}$

We conclude that $\theta =\pi$

Hence, the correct option is (D).