Question-176159

Question Number 176159 by peter frank last updated on 14/Sep/22

Answered by adhigenz last updated on 14/Sep/22

Let x = cos α + i.sin α = e^(i.α) and y = cos β + i.sin β = e^(i.β) (a). x^m y^n + (1/(x^m y^n )) = e^(i.mα) .e^(i.nβ) + e^(i.m(−α)) .e^(i.n(−β)) = e^(i(mα+nβ)) + e^(i(−mα−nβ)) = cos (mα+nβ) + i.sin (mα+nβ) + cos (−mα−nβ) + i.sin (−mα−nβ) = 2cos (mα+nβ)

$$\mathrm{Let}\:{x}\:=\:\mathrm{cos}\:\alpha\:+\:{i}.\mathrm{sin}\:\alpha\:=\:{e}^{{i}.\alpha} \:\mathrm{and}\:{y}\:=\:\mathrm{cos}\:\beta\:+\:{i}.\mathrm{sin}\:\beta\:=\:{e}^{{i}.\beta} \\ $$$$\left(\mathrm{a}\right).\:{x}^{{m}} {y}^{{n}} \:+\:\frac{\mathrm{1}}{{x}^{{m}} {y}^{{n}} }\:=\:{e}^{{i}.{m}\alpha} .{e}^{{i}.{n}\beta} \:+\:{e}^{{i}.{m}\left(−\alpha\right)} .{e}^{{i}.{n}\left(−\beta\right)} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:{e}^{{i}\left({m}\alpha+{n}\beta\right)} \:+\:{e}^{{i}\left(−{m}\alpha−{n}\beta\right)} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\mathrm{cos}\:\left({m}\alpha+{n}\beta\right)\:+\:{i}.\mathrm{sin}\:\left({m}\alpha+{n}\beta\right)\:+\:\mathrm{cos}\:\left(−{m}\alpha−{n}\beta\right)\:+\:{i}.\mathrm{sin}\:\left(−{m}\alpha−{n}\beta\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\mathrm{2cos}\:\left({m}\alpha+{n}\beta\right) \\ $$

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Question-176159

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