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Question Number 43452 by mondodotto@gmail.com last updated on 10/Sep/18

prove that tanhx=itanx

$$\mathrm{prove}\:\mathrm{that} \\ $$$$\boldsymbol{\mathrm{tanh}{x}}=\boldsymbol{{i}\mathrm{tan}{x}} \\ $$

Answered by alex041103 last updated on 10/Sep/18

tanh x = ((e^x −e^(−x) )/(e^x +e^(−x) ))=((e^(i×ix) −e^(−i×ix) )/(e^(i×ix) +e^(−i×ix) ))= =((cos ix + isin ix − cos ix + isin x)/(cos ix + isin ix + cos ix− isin x))= =i((2sin ix)/(2cos ix))=itan ix where e^(iθ) =cos iθ + isin iθ ⇒tanh x = itan ix

$$\mathrm{tanh}\:{x}\:=\:\frac{{e}^{{x}} −{e}^{−{x}} }{{e}^{{x}} +{e}^{−{x}} }=\frac{{e}^{{i}×{ix}} −{e}^{−{i}×{ix}} }{{e}^{{i}×{ix}} +{e}^{−{i}×{ix}} }= \\ $$$$=\frac{\mathrm{cos}\:{ix}\:+\:{i}\mathrm{sin}\:{ix}\:−\:\mathrm{cos}\:{ix}\:+\:{i}\mathrm{sin}\:{x}}{\mathrm{cos}\:{ix}\:+\:{i}\mathrm{sin}\:{ix}\:+\:\mathrm{cos}\:{ix}−\:{i}\mathrm{sin}\:{x}}= \\ $$$$={i}\frac{\mathrm{2sin}\:{ix}}{\mathrm{2cos}\:{ix}}={i}\mathrm{tan}\:{ix} \\ $$$${where}\:{e}^{{i}\theta} =\mathrm{cos}\:{i}\theta\:+\:{i}\mathrm{sin}\:{i}\theta \\ $$$$\Rightarrow\mathrm{tanh}\:{x}\:=\:{i}\mathrm{tan}\:{ix} \\ $$

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