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# If $x+\large\frac{1}{x}$$=2\cos\;\theta and y+\large\frac{1}{y}=$$2 \cos \;\phi$ show that $\large\frac{x^{m}}{y^{n}}+ \frac{y^{n}}{x^{m}} $$= 2 \cos \left ( m\theta -n\phi \right ) where m, n N. This is the first part of the multi-part question Q8. Can you answer this question? ## 1 Answer 0 votes Toolbox: • From De moivre's theorem we have • (i) (\cos\theta+i\sin\theta)^n=\cos n\theta+i\sin n\theta,n\in Q • (ii) (\cos\theta+i\sin\theta)^{-n}=\cos n\theta-i\sin n\theta • (iii) (\cos\theta-i\sin\theta)^n=\cos n\theta-i\sin n\theta • (iv) (\sin \theta+i\cos \theta)^n=[\cos(\large\frac{\pi}{2}$$-\theta)+i\sin(\large\frac{\pi}{2}$$-\theta)]^n=\cos n(\large\frac{\pi}{2}$$-\theta)+i\sin n(\large\frac{\pi}{2}$$-\theta) • e^{i\theta}=\cos\theta+i\sin\theta • e^{-i\theta}=\cos\theta-i\sin\theta,also written as \cos\theta and \cos(-\theta) Step 1: x+\large\frac{1}{x}$$=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 $\cos\theta-i\sin\theta=\large\frac{1}{x}$
Similarly $y+\large\frac{1}{y}$$=2\cos\phi from which have y=\cos\phi+i\sin\phi,\large\frac{1}{y}$$=\cos\phi-i\sin\phi$
Step 2:
Now $\large\frac{x^m}{y^n}+\frac{y^n}{x^m}=\frac{(\cos\theta+i\sin\theta)^m}{(\cos\phi+i\sin\phi)^n}+\frac{(\cos\phi+i\sin\phi)^n}{(\cos\theta+i\sin\theta)^m}$
$\qquad\qquad\;\;\;\;\;\;\;=(\cos\theta+i\sin\theta)^m(\cos\phi+i\sin\phi)^{-n}+(\cos\phi+i\sin\phi)^n(\cos\theta+i\sin\theta)^{-m}$
$\qquad\qquad\;\;\;\;\;\;\;=(e^{i\theta})^m.(e^{i\phi})^{-n}+(e^{i\phi})^n(e^{i\theta})^{-m}$
$\qquad\qquad\;\;\;\;\;\;\;=e^{\large im\theta-in\phi}+e^{\large in\phi-im\theta}$
$\qquad\qquad\;\;\;\;\;\;\;=e^{\large i(m\theta-n\phi)}+e^{\large -i(m\theta-n\phi)}$
$\qquad\qquad\;\;\;\;\;\;\;=\cos(m\theta-n\phi)+i\sin(m\theta-n\phi)+\cos(m\theta-n\phi)-i\sin(m\theta-n\phi)$
$\qquad\qquad\;\;\;\;\;\;\;=2\cos(m\theta-n\phi)$