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Question Number 120558 by peter frank last updated on 01/Nov/20

Answered by mr W last updated on 01/Nov/20

$$from$$$$e^{i\theta }=\cos \theta +i\sin \theta$$$$e^{-i\theta }=\cos \theta -i\sin \theta$$$$weget$$$$\cos \theta =\frac{e^{i\theta }+e^{-i\theta }}{2}$$$$with\theta =3+i:$$$$\cos (3+i)=\frac{e^{i(3+i)}+e^{-i(3+i)}}{2}$$$$=\frac{e^{3i-1}+e^{-3i+1}}{2}$$$$=\frac{1}{2e}{e}^{3i}+\frac{e}{2}{e}^{-3i}$$$$=\frac{1}{2e}(\cos 3+i\sin 3)+\frac{e}{2}(\cos 3-i\sin 3)$$$$=\frac{\cos 3}{2}(e+\frac{1}{e})-\frac{\sin 3}{2}(e-\frac{1}{e})i$$

Commented by peter frank last updated on 01/Nov/20

$$\mathrm{thank}\mathrm{you}$$

Answered by TANMAY PANACEA last updated on 01/Nov/20

$$cos3cosi-sin3sini$$$$cos3cosh-sin3\times isinh$$

Answered by mathmax by abdo last updated on 01/Nov/20

$$\cos (3+\mathrm{i})=\frac{{\mathrm{e}}^{\mathrm{i}(3+\mathrm{i})}+{\mathrm{e}}^{-\mathrm{i}(3+\mathrm{i})}}{2}=\frac{{\mathrm{e}}^{3\mathrm{i}-1}+{\mathrm{e}}^{-3\mathrm{i}+1}}{2}$$$$=\frac{1}{2}\{{\mathrm{e}}^{-1}(\cos \left(3\right)+\mathrm{isin}3)+\mathrm{e}(\cos 3-\mathrm{isin}3)\}$$$$=\frac{1}{2}\{(\mathrm{e}+{\mathrm{e}}^{-1})\cos 3+\mathrm{isin}3({\mathrm{e}}^{-1}-\mathrm{e})\}$$$$=\frac{\mathrm{e}+{\mathrm{e}}^{-1}}{2}\cos \left(3\right)-\mathrm{isin}\left(3\right)\left(\frac{\mathrm{e}-{\mathrm{e}}^{-1}}{2}\right)=\mathrm{ch}\left(1\right)\cos \left(3\right)-\mathrm{ish}\left(1\right)\sin 3$$

Commented by peter frank last updated on 01/Nov/20

$$\mathrm{thank}\mathrm{you}$$

Commented by mathmax by abdo last updated on 08/Nov/20

$$\mathrm{you}\mathrm{are}\mathrm{welcome}$$

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