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Q)

If $x\;=\cos\;\alpha +i\sin\;\alpha\;;\; y\;= \cos\;\beta + i\sin \;\beta $ prove that $x^{m}y^{n} \;+\large \frac{1}{x^{m}y^{n}}$$= \;2\cos \;\left ( m\alpha + n\beta \right )$

where  m, n N.

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A)
Toolbox:
  • From De moivre's theorem we have
  • (i) $(\cos\theta+i\sin\theta)^n=\cos n\theta+i\sin n\theta,n\in Q$
  • (ii) $(\cos\theta+i\sin\theta)^{-n}=\cos n\theta-i\sin n\theta$
  • (iii) $(\cos\theta-i\sin\theta)^n=\cos n\theta-i\sin n\theta$
  • (iv) $(\sin \theta+i\cos \theta)^n=[\cos(\large\frac{\pi}{2}$$-\theta)+i\sin(\large\frac{\pi}{2}$$-\theta)]^n=\cos n(\large\frac{\pi}{2}$$-\theta)+i\sin n(\large\frac{\pi}{2}$$-\theta)$
  • $e^{i\theta}=\cos\theta+i\sin\theta$
  • $e^{-i\theta}=\cos\theta-i\sin\theta$,also written as $\cos\theta$ and $\cos(-\theta)$
Step 1:
$x=\cos\alpha+i\sin\alpha$
$y=\cos\beta+i\sin\beta$
$\Rightarrow x=e^{\large i\alpha}$
$y=e^{\large i\beta}$
$x^my^n+\large\frac{1}{x^my^n}$$=(e^{i\alpha})^m(e^{i\beta})^n+\large\frac{1}{(e^{i\alpha})^m(e^{i\beta})}$
$\qquad\qquad\;\;\;=e^{im\alpha}e^{im\beta}+\large\frac{1}{e^{\Large im\alpha}e^{\Large im\beta}}$
$\qquad\qquad\;\;\;=e^{\large i(m\alpha+n\beta)}+e^{\large-i(m\alpha+n\beta)}$
$\qquad\qquad\;\;\;=\cos(m\alpha+n\beta)+i\sin(m\alpha+n\beta)+(\cos(m\alpha+n\beta)-i\sin(m\alpha+\beta))$
$\qquad\qquad\;\;\;=2\cos(m\alpha+n\beta)$
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