simplifying $-\pi i/8 (e^{i\pi/8} + e^{i3\pi/8} + e^{i5\pi/8} + e^{i7\pi/8})$

Question

simplifying $-\pi i/8 (e^{i\pi/8} + e^{i3\pi/8} + e^{i5\pi/8} + e^{i7\pi/8})$

in my lecture notes somehow my lecture got from$-\pi i/8 (e^{i\pi/8} + e^{i3\pi/8} + e^{i5\pi/8} + e^{i7\pi/8})$ to $\dfrac{\pi}{8\sin(\pi/8)}$, could someone show me how? I've spent the last 90 minutes on it!

Answer 1

$$-\frac{\pi i}{8} (e^{\pi i/8} + e^{3\pi i/8} + e^{5\pi i/8} + e^{ 7\pi i/8})=-\frac{e^{\pi i/8}\pi i}8\left(1+e^{\pi i/4}+e^{2\pi i/4}+e^{3\pi i/4}\right)=$$

$$=-\frac{e^{\pi i/8}\pi i}8\frac{\overbrace{\left(e^{\pi i/4}\right)^4-1}^{=-2}}{\underbrace{e^{\pi i/4}-1}_{=\frac1{\sqrt2}(1+i)-1=}}=\frac{e^{\pi i/8}\pi i}{4\left(\frac1{\sqrt2}+\frac i{\sqrt2}-1\right)}=\ldots$$

I'll stop here as, if there's no mistake, the above number isn't real, as you say someone got.

Answer 2

Recall $e^{i\theta} = \cos\theta + i\sin\theta$. We have, then $\cos\pi/8 + \cos3\pi/8 + cos5\pi/8 + cos7\pi/8$ which cancels out to zero because $\cos\pi/8=-\cos7\pi/8$ and similarly with the other angles, and for the sins we have $2\sin\pi/8 + 2\sin3\pi/8$, so the answer would be $\pi/8(2\sin\pi/8 + 2\sin3\pi/8)$. Note that this doesn't simplify to what you want, but WolframAlpha says that the original simplification is wrong.

Answer 3

Trying out $-i=e^{-\frac{1}{2}\pi i}$ I find $$\cdots=\frac{\pi}{8}\left(e^{-\frac{3}{8}\pi i}+e^{-\frac{1}{8}\pi i}+e^{\frac{1}{8}\pi i}+e^{\frac{3}{8}\pi i}\right)=\frac{\pi}{8}\left(2\cos\frac{3}{8}\pi+2\cos\frac{1}{8}\pi\right)$$