real part of cos(1+iota)

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Please find below the solution to the asked query:

$$\begin{gathered} \mathrm{We} \mathrm{know} \mathrm{that} \\ {\mathrm{e}}^{\mathrm{i\theta }}=\mathrm{cos\theta }+\mathrm{isin\theta } \\ \mathrm{Let} \mathrm{z}=1+\mathrm{i} \\ \mathrm{Hence} \mathrm{cosz} \mathrm{can} \mathrm{be} \mathrm{rewritten} \mathrm{as} \\ \cos \left(\mathrm{z}\right)=\frac{{\mathrm{e}}^{\mathrm{iz}}+{\mathrm{e}}^{-\mathrm{iz}}}{2} \\ =\frac{{\mathrm{e}}^{\mathrm{i}\left(1+\mathrm{i}\right)}+{\mathrm{e}}^{-\mathrm{i}\left(1+\mathrm{i}\right)}}{2} \\ =\frac{{\mathrm{e}}^{\mathrm{i}}.{\mathrm{e}}^{{\mathrm{i}}^2}+{\mathrm{e}}^{-\mathrm{i}}.{\mathrm{e}}^{-{\mathrm{i}}^2}}{2} \\ =\frac{{\mathrm{e}}^{\mathrm{i}}.{\mathrm{e}}^{-1}+{\mathrm{e}}^{-\mathrm{i}}.{\mathrm{e}}^1}{2} \\ =\frac{{\mathrm{e}}^{-1}\left(\cos 1+\mathrm{isin}1\right)+\left(\cos 1-\mathrm{isin}1\right).{\mathrm{e}}^1}{2} \\ =\frac{{\displaystyle \frac{1}{\mathrm{e}}\left(\cos 1+\mathrm{isin}1\right)+\left(\cos 1-\mathrm{isin}1\right).\mathrm{e}}}{2} \\ \Rightarrow \cos \left(\mathrm{z}\right)=\frac{\cos 1\left({\displaystyle \frac{1}{\mathrm{e}}+\mathrm{e}}\right)+\sin 1\left({\displaystyle \frac{1}{\mathrm{e}}-\mathrm{e}}\right)}{2} \\ \Rightarrow \cos \left(1+\mathrm{i}\right)=\frac{\cos 1\left({\displaystyle \frac{1}{\mathrm{e}}+\mathrm{e}}\right)+\sin 1\left({\displaystyle \frac{1}{\mathrm{e}}-\mathrm{e}}\right)}{2} \\ \mathbf{Hence}\mathbf{ }\mathbf{real}\mathbf{ }\mathbf{part}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{cos}\left(\mathbf{1}\mathbf{+}\mathbf{i}\right)\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{cos}\mathbf{1}\left(\frac{\mathbf{1}}{\mathbf{e}}\mathbf{+}\mathbf{e}\right)\mathbf{+}\mathbf{ }\left[\mathbf{Answer}\right] \end{gathered}$$

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