Question

If \[ e^{i\theta}=\operatorname{cis}\theta, \] then \[ \sum_{n=0}^{\infty}\frac{\cos(n\theta)}{2^n} = \]

• A. \(\dfrac{4+2\cos\theta}{5-4\cos\theta}\)
• B. \(\dfrac{4-2\cos\theta}{5+4\cos\theta}\)
• C. \(\dfrac{4-2\cos\theta}{5-4\cos\theta}\)
• D. \(\dfrac{4+2\cos\theta}{5+4\cos\theta}\)

Hint:

For series involving \(\cos(n\theta)\), use \[ \cos(n\theta)=\operatorname{Re}(e^{in\theta}) \] and convert the series into a geometric series.

Answer 1

The Correct Option is C

Step 1: Express \(\cos(n\theta)\) using complex numbers.
Using Euler's formula, \[ e^{i\theta}=\cos\theta+i\sin\theta \] Also, \[ e^{in\theta}=\cos(n\theta)+i\sin(n\theta) \] Therefore, \[ \cos(n\theta)=\operatorname{Re}(e^{in\theta}) \]

Step 2: Convert the given series into a geometric series.
Let \[ S=\sum_{n=0}^{\infty}\frac{\cos(n\theta)}{2^n} \] Then, \[ S=\operatorname{Re}\left(\sum_{n=0}^{\infty}\frac{e^{in\theta}}{2^n}\right) \] Now, \[ \sum_{n=0}^{\infty}\frac{e^{in\theta}}{2^n} = \sum_{n=0}^{\infty}\left(\frac{e^{i\theta}}{2}\right)^n \] This is an infinite geometric series with common ratio \[ r=\frac{e^{i\theta}}{2} \] Since \[ |r|=\frac{1}{2}\lt 1, \] the series converges.

Step 3: Sum the geometric series.
Using \[ \sum_{n=0}^{\infty}r^n=\frac{1}{1-r}, \] we get \[ \sum_{n=0}^{\infty}\left(\frac{e^{i\theta}}{2}\right)^n = \frac{1}{1-\frac{e^{i\theta}}{2}} \] \[ = \frac{2}{2-e^{i\theta}} \] Therefore, \[ S=\operatorname{Re}\left(\frac{2}{2-e^{i\theta}}\right) \]

Step 4: Rationalize the complex denominator.
Since \[ e^{i\theta}=\cos\theta+i\sin\theta, \] we have \[ 2-e^{i\theta} = 2-\cos\theta-i\sin\theta \] Thus, \[ \frac{2}{2-e^{i\theta}} = \frac{2}{2-\cos\theta-i\sin\theta} \] Multiply numerator and denominator by the conjugate: \[ 2-\cos\theta+i\sin\theta \] So, \[ \frac{2}{2-e^{i\theta}} = \frac{2(2-\cos\theta+i\sin\theta)} {(2-\cos\theta)^2+\sin^2\theta} \] The denominator becomes \[ (2-\cos\theta)^2+\sin^2\theta \] \[ =4-4\cos\theta+\cos^2\theta+\sin^2\theta \] Using \[ \sin^2\theta+\cos^2\theta=1, \] we get \[ =5-4\cos\theta \]

Step 5: Take the real part.
Therefore, \[ \frac{2}{2-e^{i\theta}} = \frac{2(2-\cos\theta+i\sin\theta)} {5-4\cos\theta} \] The real part is \[ S= \frac{2(2-\cos\theta)} {5-4\cos\theta} \] \[ = \frac{4-2\cos\theta} {5-4\cos\theta} \]

Step 6: Final conclusion.
Hence, \[ \boxed{\frac{4-2\cos\theta}{5-4\cos\theta}} \] which corresponds to option (3).