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.
$\endgroup$ 34 Answers
$\begingroup$Note that $$\cos(\pi/12)-i\sin(\pi/12)=\cos(-\pi/12)+i\sin(-\pi/12)$$
$\endgroup$ $\begingroup$$$(\cos y-i\sin y)^n=\dfrac1{(\cos y+i\sin y)^n}=\dfrac1{\cos(ny)+i\sin(ny)}=\cos(ny)-i\sin(ny)$$
as $(\cos x+i\sin x)(\cos x-i\sin x)=1$
$\endgroup$ $\begingroup$$(\cos y-i\sin y)^n=(\cos(-y)+i\sin(-y))^n=\cos(n(-y))+i\sin(n(-y))=\cos(ny)-i\sin(ny)$
$\endgroup$ $\begingroup$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}
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