First slide

De-moivre's theorem

Question

$2 \cos A = x + \frac{1}{x}, 2 \cos B = y + \frac{1}{y} .$ Then find the value of k if $k \cos (A - B) = \frac{x}{y} + \frac{y}{x}$

Moderate

A

0
B

-1
C

2
D

4

Solution

$G i v e n 2 \cos A = x + \frac{1}{x} \Rightarrow x^{2} - 2 x \cos A + 1 = 0 \Rightarrow x = \frac{2 \cos A \pm \sqrt{4 \cos^{2} A - 4}}{2} = \frac{2 \cos A \pm i 2 \sin A}{2} = \cos A \pm i \sin A$
$S i m i l a r l y 2 \cos B = y + \frac{1}{y} \Rightarrow y = \cos B \pm i \sin B N o w \frac{x}{y} = \frac{\cos A \pm i \sin A}{\cos B \pm i \sin B} = \frac{e^{\pm i A}}{e^{\pm i B}} = e^{\pm i (A - B)}$
$S i m i l a r l y \frac{y}{x} = e^{\pm i (B - A)} ∴ \frac{x}{y} + \frac{y}{x} = e^{\pm i (A - B)} + e^{\pm i (B - A)} = \cos (A - B) + \cos (B - A) = \cos (A - B) + \cos (A - B) = 2 \cos (A - B)$
$\Rightarrow k = 2$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App