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If $\cos\alpha+\cos\beta+\cos\gamma=\sin\alpha+\sin\beta+\sin\gamma=0$ and $\cos3\alpha=\dfrac34$ then $\cos^8\alpha+\cos^8\beta+\cos^8\gamma=$

  • A) $\dfrac3{128}$
  • B) $\dfrac{27}{32}$
  • C) $\dfrac3{16}$
  • D) $\dfrac5{128}$

ATTEMPT $1$: $e^{i\alpha}=\cos\alpha+i\sin\alpha$, $e^{i\beta}=\cos\beta+i\sin\beta$, $e^{i\gamma}=\cos\gamma+i\sin\gamma$

Adding, we get $e^{i\alpha}+e^{i\beta}+e^{i\gamma}=0$

Now, if $a+b+c=0\implies a^3+b^3+c^3=3abc$

Therefore, $e^{i3\alpha}+e^{i3\beta}+e^{i3\gamma}=3e^{i(\alpha+\beta+\gamma)}$

$\implies \cos3\alpha+i\sin3\alpha+\cos3\beta+i\sin3\beta+\cos3\gamma+i\sin3\gamma=3\cos(\alpha+\beta+\gamma)+3i\sin(\alpha+\beta+\gamma)$

Sure, $\cos3\alpha$ is given but what about the rest? Also, we need $\cos^8\alpha$. How to get that? On RHS, can we use the following formula?

$\cos(A+B+C)=\cos A\cos B\cos C-\cos A\sin B \sin C-\cos B\sin C\sin A-\cos C\sin A\sin B$

ATTEMPT $2$: $$\cos3\alpha=\frac34\\4\cos^3\alpha-3\cos\alpha=\frac34\\\cos\alpha(4\cos^2\alpha-3)=\frac34\\4\cos^2\alpha-3=\frac34\sec\alpha$$

Also, $0\le4\cos^2\alpha\le4\implies-3\le4\cos^2\alpha-3\le1\implies-3\le\frac34\sec\alpha\le1\implies-4\le\sec\alpha\le\frac43$

But $\sec\alpha\ge1\implies1\le\sec\alpha\le\frac43$

Does that help?

ATTEMPT $3$: Hit and try. If $\beta=\frac{\pi}2\implies\alpha+\gamma=\pi$

But it doesn't fit sine equation. Have tried with $0,\pi$ etc. Not working.

aarbee
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    Hint: Geometrically, if 3 unit vectors in $\mathbb{R}^2$ sum to zero, then they are at $120^\circ$ with respect to each other. This implies $\beta,\gamma = \alpha \pm \frac{2\pi}{3} \pmod {2\pi}$. As a result, $e^{in\alpha} + e^{in\beta} + e^{in\gamma} = 0$ for integer $n$ which are not divisible by $3$. – achille hui Jun 14 '21 at 18:42
  • See https://math.stackexchange.com/questions/1397066/clarification-regarding-a-question – lab bhattacharjee Jun 16 '21 at 05:12

1 Answers1

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$e^{i\alpha}$, $e^{i\beta}$ and $e^{i\gamma}$ represent the three vertices of an equilateral triangle inscribed in the unit circle. WLOG, let $\beta=\alpha+\dfrac{2\pi}3$ and $\gamma=\alpha+\dfrac{4\pi}3$.

\begin{align*} \cos^8\alpha+\cos^8\beta+\cos^8\gamma&= \dfrac{(e^{i\alpha}+e^{-i\alpha})^8}{256}+\dfrac{(e^{i\beta}+e^{-i\beta})^8}{256}+\dfrac{(e^{i\gamma}+e^{-i\gamma})^8}{256} \end{align*}

Note that $e^{ik\alpha}+e^{ik\beta}+e^{ik\gamma}=0$ if $k$ is not a multiple of $3$.

Using the binomial theorem,

\begin{align*} &\;\cos^8\alpha+\cos^8\beta+\cos^8\gamma\\ =&\;\dfrac{(e^{i\alpha}+e^{-i\alpha})^8}{256}+\dfrac{(e^{i\beta}+e^{-i\beta})^8}{256}+\dfrac{(e^{i\gamma}+e^{-i\gamma})^8}{256}\\ =&\;\frac{8e^{6i\alpha}+8e^{6i\beta}+8e^{6i\gamma}+8e^{-6i\alpha}+3(70)+8e^{-6i\beta}+8e^{-6i\gamma}}{256}\\ =&\;\frac{24e^{6i\alpha}+24e^{-6i\alpha}+210}{256}\\ =&\;\frac{48\cos6\alpha+210}{256}\\ =&\;\frac{48(2\cos^23\alpha-1)+210}{256}\\ =&\;\frac{27}{32} \end{align*}

CY Aries
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