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Question Number 51897 by Tawa1 last updated on 31/Dec/18

$$\mathrm{If}\mathrm{x}+\frac{1}{\mathrm{x}}=2\cos \theta ,\mathrm{y}+\frac{1}{\mathrm{y}}=2\cos \varphi ,\mathrm{z}+\frac{1}{\mathrm{z}}=2\cos \psi$$$$\mathrm{Show}\mathrm{that}\mathrm{xyz}+\frac{1}{\mathrm{xyz}}=2\cos (\theta +\varphi +\psi )$$

Answered by behi83417@gmail.com last updated on 31/Dec/18

$$cos\theta =\frac{e^{i\theta }+e^{-i\theta }}{2}=\frac{1}{2}(e^{i\theta }+\frac{1}{e^{i\theta }})=\frac{1}{2}(x+\frac{1}{x})$$$$wecanput:x=e^{i\theta },y=e^{i\mathrm{\varnothing }},z=e^{i\psi }$$$$\Rightarrow xyz+\frac{1}{xyz}=e^{i\theta +i\varphi +i\psi }+\frac{1}{e^{i\theta +i\varphi +i\psi }}=$$$$=e^{i(\theta +\varphi +\psi )}+\frac{1}{e^{i(\theta +\varphi +\psi )}}=2cos(\theta +\varphi +\psi ).$$

Commented by Tawa1 last updated on 01/Jan/19

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}$$

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