# If $x\;=\cos\;\alpha +i\sin\;\alpha\;;\; y\;= \cos\;\beta + i\sin \;\beta$ prove that $x^{m}y^{n} \;+\large \frac{1}{x^{m}y^{n}}$$= \;2\cos \;\left ( m\alpha + n\beta \right ) where m, n N. ## 1 Answer Comment A) Toolbox: • From De moivre's theorem we have • (i) (\cos\theta+i\sin\theta)^n=\cos n\theta+i\sin n\theta,n\in Q • (ii) (\cos\theta+i\sin\theta)^{-n}=\cos n\theta-i\sin n\theta • (iii) (\cos\theta-i\sin\theta)^n=\cos n\theta-i\sin n\theta • (iv) (\sin \theta+i\cos \theta)^n=[\cos(\large\frac{\pi}{2}$$-\theta)+i\sin(\large\frac{\pi}{2}$$-\theta)]^n=\cos n(\large\frac{\pi}{2}$$-\theta)+i\sin n(\large\frac{\pi}{2}$$-\theta) • e^{i\theta}=\cos\theta+i\sin\theta • e^{-i\theta}=\cos\theta-i\sin\theta,also written as \cos\theta and \cos(-\theta) Step 1: x=\cos\alpha+i\sin\alpha y=\cos\beta+i\sin\beta \Rightarrow x=e^{\large i\alpha} y=e^{\large i\beta} x^my^n+\large\frac{1}{x^my^n}$$=(e^{i\alpha})^m(e^{i\beta})^n+\large\frac{1}{(e^{i\alpha})^m(e^{i\beta})}$
$\qquad\qquad\;\;\;=e^{im\alpha}e^{im\beta}+\large\frac{1}{e^{\Large im\alpha}e^{\Large im\beta}}$
$\qquad\qquad\;\;\;=e^{\large i(m\alpha+n\beta)}+e^{\large-i(m\alpha+n\beta)}$
$\qquad\qquad\;\;\;=\cos(m\alpha+n\beta)+i\sin(m\alpha+n\beta)+(\cos(m\alpha+n\beta)-i\sin(m\alpha+\beta))$
$\qquad\qquad\;\;\;=2\cos(m\alpha+n\beta)$