Question

If $2\cos \theta =x+\frac{1}{x}$ and $2\cos \varphi =y+\frac{1}{y}$, then

• A. $x^n+\frac{1}{x^n}=2\cos \left(n\theta \right)$
• B. $\frac{x}{y}+\frac{y}{x}=2\cos (\theta -\varphi )$
• C. $xy+\frac{1}{xy}=2\cos (\theta +\varphi )$
• D. None of these

Answer 1

Answer: (A), (B), (C)

The correct options are
A $x^n+\frac{1}{x^n}=2\cos \left(n\theta \right)$
B $\frac{x}{y}+\frac{y}{x}=2\cos (\theta -\varphi )$
C $xy+\frac{1}{xy}=2\cos (\theta +\varphi )$

Considering $x=cos\theta +isin\theta$
$=e^{i\theta }$
Therefore
$\overline{x}=\frac{1}{x}=e^{-i\theta }$
$x+\frac{1}{x}=2cos\theta$
$x^n=e^{in\theta }=cosn\theta +isinn\theta$
$\frac{1}{x^n}=e^{-in\theta }$
$=cosn\theta -isinn\theta$
Hence $x^n+\frac{1}{x^n}$
$=2cosn\theta$
Similarly considering $y=cos\varphi +isin\varphi$
$=e^{i\varphi }$
Therefore
$\overline{y}=\frac{1}{y}=e^{-i\varphi }$
$y+\frac{1}{y}=2cos\varphi$
$y^n=e^{in\varphi }=cosn\varphi +isinn\varphi$
$\frac{1}{y^n}=e^{-in\varphi }$
$=cosn\varphi -isinn\varphi$
Hence $y^n+\frac{1}{y^n}$
$=2cosn\varphi$
$xy+\frac{1}{xy}$
$e^{i(\theta +\varphi )}+e^{-i(\theta +\varphi )}$
$=2cos(\theta +\varphi )$
$x{y}^{-1}+y{x}^{-1}$
$=e^{i(\theta -\varphi )}+e^{-i(\theta -\varphi )}$
$=2cos(\theta -\varphi )$
Hence, options 'B', 'C' and 'D' are correct.