Question and Answers Forum

All Questions      Topic List

None Questions

Previous in All Question      Next in All Question      

Previous in None      Next in None      

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

$$\mathrm{prove}\mathrm{that}$$$$\mathbf{tanh}\mathit{x}=\mathit{i}\mathbf{tan}\mathit{x}$$

Answered by alex041103 last updated on 10/Sep/18

$$\tanh x=\frac{e^x-e^{-x}}{e^x+e^{-x}}=\frac{e^{i\times ix}-e^{-i\times ix}}{e^{i\times ix}+e^{-i\times ix}}=$$$$=\frac{\cos ix+i\sin ix-\cos ix+i\sin x}{\cos ix+i\sin ix+\cos ix-i\sin x}=$$$$=i\frac{2\sin ix}{2\cos ix}=i\tan ix$$$$where{e}^{i\theta }=\cos i\theta +i\sin i\theta$$$$\Rightarrow \tanh x=i\tan ix$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com