logo

Ask Questions, Get Answers

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

Prove that $\left ( 1+ \cos \; \theta + \mathit{i}\sin \; \theta \right )^{\mathit{n}} + \left ( 1+ \cos \; \theta - \mathit{i}sin \; \theta \right )^{\mathit{n}} = 2^{\mathit{n+1}} \cos ^{\mathit{n}}\left ( \large\frac{\theta}{2} \right )$$ \cos \;\large\frac{n\theta }{2}$

m, n N.

This is the third part of the multi-part question Q4.

Can you answer this question?
 
 

1 Answer

0 votes
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:
Now $(1+\cos\theta+i\sin\theta)=2\cos^2\large\frac{\theta}{2}$$+2i\sin\large\frac{\theta}{2}$$\cos\large\frac{\theta}{2}$
$\qquad\qquad\;\;\;\;\;\qquad\qquad=2\cos\large\frac{\theta}{2}$$(\cos\large\frac{\theta}{2}$$+i\sin\large\frac{\theta}{2})$
$(1+\cos\theta-i\sin\theta)=(1+\cos\theta+i\sin\theta)=2\cos\large\frac{\theta}{2}$$(\cos\large\frac{\theta}{2}$$-i\sin\large\frac{\theta}{2})$
Step 2:
LHS=$[2\cos\large\frac{\theta}{2}$$(\cos\large\frac{\theta}{2}$$+i\sin\large\frac{\theta}{2}]^n+$$[2\cos\large\frac{\theta}{2}($$\cos\large\frac{\theta}{2}$$-i\sin\large\frac{\theta}{2}]^n$
$\quad=2^n\cos^n\large\frac{\theta}{2}$$[\cos\large\frac{n\theta}{2}+$$i\sin\large\frac{n\theta}{2}+$$\cos\large\frac{n\theta}{2}$$-i\sin\large\frac{n\theta}{2}]$
$\quad=2^n\cos^n\large\frac{\theta}{2}$$.2\cos\large\frac{n\theta}{2}$
$\quad=2^{n+1}\cos^n\large\frac{\theta}{2}$$.\cos\large\frac{n\theta}{2}$
$\quad$=RHS.
Hence proved.
answered Jun 11, 2013 by sreemathi.v
 

Related questions

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