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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.

user402016
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  • Use https://math.stackexchange.com/questions/3510/how-to-prove-eulers-formula-ei-varphi-cos-varphi-i-sin-varphi – lab bhattacharjee Aug 26 '18 at 11:45
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    By the way, you're getting the cosine and sine wrong values in the last part of your computation. It should be $-\frac12 - \frac{\sqrt3}2i$. – hmakholm left over Monica Aug 26 '18 at 11:53
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    I can use De Moivre's formula for the left part Call that $z$, then the right part is just $\bar z$, so their sum is $z + \bar z = 2 \operatorname{Re}(z)$. But you need to fix the left part, first. – dxiv Aug 27 '18 at 05:25

4 Answers4

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

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

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

Angina Seng
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lab bhattacharjee
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$(\cos y-i\sin y)^n=(\cos(-y)+i\sin(-y))^n=\cos(n(-y))+i\sin(n(-y))=\cos(ny)-i\sin(ny)$

drhab
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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}

Bernard
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