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# 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.