Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
Home  >>  TN XII Math  >>  Complex Numbers
0 votes

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.



Can you answer this question?

1 Answer

0 votes
  • 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

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, AIPMT Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App