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If $x +\large\frac{1}{x} = $$2\; cos\;\theta $ prove that $x^{n} -\large \frac{1}{x^{n}}= $$2i\; sin\;n\theta $

where n N.

This is the second part of the multi-part question Q7.



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1 Answer

  • 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:
$\qquad\quad\;=\cos n\theta+i\sin n\theta-(\cos n\theta-i\sin n\theta)$
$\qquad\quad\;=2i\sin n\theta$
Hence proved.
answered Jun 12, 2013 by sreemathi.v

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