If $ x+\dfrac{1}{x}=2\cos \theta \ and \ y+\dfrac{1}{y}=2\cos \phi$ then which of the following is not correct?
- $\displaystyle \frac{x}{y} +\frac{y}{x}=2\cos \left ( \theta -\phi \right )$
- $x^{m}y^{n}=\cos \left ( m\theta +n\phi \right )+i\sin \left ( m\theta +n\phi \right )$
- $x^{m}y^{n}+x^{-m}y^{-n}=2\cos \left ( m\theta +n\phi \right )$
-
None of these
$x+\dfrac { 1 }{ x } =2\cos { \theta } \quad & \quad y+\dfrac { 1 }{ y } =2\cos { \phi } \ $
$\Rightarrow x=\cos { \theta } +i\sin { \theta } =cis\theta \ \quad & \quad y=\cos { \phi } +i\sin { \phi } =cis\phi \ $
$\dfrac { x }{ y } +\dfrac { y }{ x } =\dfrac { cis\theta }{ cis\phi } +\dfrac { cis\phi }{ cis\theta } =cis\left( \theta -\phi \right) +cis\left( -\theta +\phi \right) $
$\therefore \quad \dfrac { x }{ y } +\dfrac { y }{ x } =2\cos { \left( \theta -\phi \right) } $
${ x }^{ m }{ y }^{ n }={ \left( cis\theta \right) }^{ m }{ \left( cis\phi \right) }^{ n }=\left( cism\theta \right) \left( cisn\phi \right) $ ...{ De Moivre's Theorem}
$\therefore \quad { x }^{ m }{ y }^{ n }=cis\left( m\theta +n\phi \right) =\cos { \left( m\theta +n\phi \right) } +i\sin { \left( m\theta +n\phi \right) } $
${ x }^{ -m }{ y }^{ -n }={ \left( cis\theta \right) }^{ -m }{ \left( cis\phi \right) }^{ -n }=\left( cis\left( -m\theta \right) \right) \left( cis\left( -n\phi \right) \right) \ \therefore \quad { x }^{ -m }{ y }^{ -n }=cis\left( -m\theta -n\phi \right) =\cos { \left( m\theta +n\phi \right) } -i\sin { \left( m\theta +n\phi \right) } \ $
$\therefore \quad { x }^{ m }{ y }^{ n }+{ x }^{ -m }{ y }^{ -n }=2\cos { \left( m\theta +n\phi \right) } $