Complex Numbers and Hyperbolic Functions

Question

How would you evaluate: $\mathfrak{R}\left[(1+i)\sin\left(\dfrac{(2+i)\pi}{4}\right)\right]$?

I know that $\cos x = \dfrac{e^{ix}+e^{-ix}}{2}$ and $\sin x = \dfrac{e^{ix}-e^{-ix}}{2i}$.

I have also tried expanding the $\sin\left(\dfrac{(2+i)\pi}{4}\right)$ part to $\sin\left(\dfrac{\pi}{2}\right)\cos\left(\dfrac{i\pi}{4}\right)+\cos\left(\dfrac{\pi}{2}\right)\sin\left(\dfrac{i\pi}{4}\right)$ using the addition formula for $\sin$ angle.

Answer 1

We know that $$ \sin\left(\dfrac{(2+i)\pi}{4}\right)=\sin\left(\dfrac{\pi}{2}+\frac{i\pi}{4}\right)=\cos\left(\frac{i\pi}{4}\right), $$ then $$ \begin{align} (1+i)\sin\left(\dfrac{(2+i)\pi}{4}\right)&= (1+i)\cos\left(\frac{i\pi}{4}\right)\\ &=(1+i)\left(\dfrac{e^{i\frac{i\pi}{4}}+e^{-i\frac{i\pi}{4}}}{2}\right)\\ &=(1+i)\left(\dfrac{e^{-\frac{\pi}{4}}+e^{\frac{\pi}{4}}}{2}\right)\\ &=\left(\dfrac{e^{-\frac{\pi}{4}}+e^{\frac{\pi}{4}}}{2}\right)+i\left(\dfrac{e^{-\frac{\pi}{4}}+e^{\frac{\pi}{4}}}{2}\right). \end{align} $$

Answer 2

$$\sin\frac{(2+i)\pi}4=\sin\left(\frac\pi2+\frac\pi4i\right)=\cos\frac\pi4i$$

But

$$\cos\frac\pi4i=\frac12\left(e^{-\frac\pi4}+e^{\frac\pi4}\right)\implies$$

$$(1+i)\sin\frac{(2+i)\pi}4=\ldots$$