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Home  >>  TN XII Math  >>  Complex Numbers
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If $\; x+\large\frac{1}{x}= $$\;2\;\cos\;\theta \; $ prove that $\;x^{n}+\large\frac{1}{x^{n}}$$= \;2\;\cos\;n\theta\; $

This is the first part of the multi-part Q7.

 

where n N

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1 Answer

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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+\large\frac{1}{x}$$=2\cos \theta$
$\Rightarrow x^2-2x\cos\theta+1=0$
Solving the equations $x=\large\frac{2\cos\theta+\sqrt{4\cos^2\theta-4}}{2}$
$\Rightarrow \large\frac{2\cos\theta\pm 2i\sin\theta}{2}$
$\Rightarrow \cos\theta\pm i\sin\theta$
Let one root be given by $x=\cos\theta+i\sin\theta$
The other root is $\cos\theta-i\sin\theta=\large\frac{1}{x}$
Step 2:
$x^n+\large\frac{1}{x^n}$$=(\cos\theta+i\sin\theta)^n+(\cos\theta-i\sin\theta)^n$
$\qquad\quad\;=\cos n\theta+i\sin n\theta+\cos n\theta-i\sin n\theta$
$\qquad\quad\;=2\cos n\theta$
Hence proved.
answered Jun 11, 2013 by sreemathi.v
 

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