What is $[\cos(\pi/12)+i\sin(\pi/12)]^{16}+[\cos(\pi/12)-i\sin(\pi/12)]^{16}$?

Question

What is $[\cos(\pi/12)+i\sin(\pi/12)]^{16}+[\cos(\pi/12)-i\sin(\pi/12)]^{16}$?

I can use De Moivre's formula for the left part:

$[\cos(\pi/12)+i\sin(\pi/12)]^{16} = \cos(4\pi/3) + i\sin(4\pi/3) = -\dfrac{\sqrt3}{2} + \dfrac{i}{2}$

but I'm stuck at the right part. Thanks in advance.

Answer 1

Note that $$\cos(\pi/12)-i\sin(\pi/12)=\cos(-\pi/12)+i\sin(-\pi/12)$$

Answer 2

$$(\cos y-i\sin y)^n=\dfrac1{(\cos y+i\sin y)^n}=\dfrac1{\cos(ny)+i\sin(ny)}=\cos(ny)-i\sin(ny)$$

as $(\cos x+i\sin x)(\cos x-i\sin x)=1$

Answer 3

$(\cos y-i\sin y)^n=(\cos(-y)+i\sin(-y))^n=\cos(n(-y))+i\sin(n(-y))=\cos(ny)-i\sin(ny)$

Answer 4

It's much simpler with the exponential notation: \begin{align} (\cos\pi/12+i\sin\pi/12) ^{16}&+(\cos\pi/12-i\sin\pi/12 )^{16}\\&=\mathrm e^{\tfrac{4i\pi}3}+\mathrm e^{\tfrac{-4i\pi}3}=2\cos\frac{4\pi}3=2\cdot\frac12 \end{align}