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.

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$
$\qquad\quad\;=\cos n\theta+i\sin n\theta-(\cos n\theta-i\sin n\theta)$
$\qquad\quad\;=2i\sin n\theta$
Hence proved.