MHB Tricky Trigonometry: Evaluating Cosine Cubes Without a Calculator

[anemone]

Here is this week's POTW:

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Without using a calculator, evaluate $\cos^3 \left(\dfrac{2\pi}{7}\right)+\cos^3 \left(\dfrac{4\pi}{7}\right)+\cos^3 \left(\dfrac{8\pi}{7}\right)$.

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[anemone]

Congratulations to the following members for their correct solution: (Smile)

1. castor28
2. greg1313
3. lfdahl
4. kaliprasad

Solution from kaliprasad:

Spoiler

We have $4\cos^3 x = \cos\, x + 3 \cos 3x$ and by using the formula $\cos(x) =\cos(2\pi-x) = \cos(4\pi-x)$, we get

$4(\cos^3\dfrac{2\pi}{7} + \cos^3\dfrac{4\pi}{7} + \cos^3\dfrac{8\pi}{7})$
=$\cos\dfrac{2\pi}{7} +3\cos\dfrac{6\pi}{7} + \cos\dfrac{4\pi}{7} +3\cos\dfrac{12\pi}{7} + \cos\dfrac{8\pi}{7} +3\cos\dfrac{24\pi}{7}$
$$=(\cos\frac{2\pi}{7} +\cos\frac{4\pi}{7} + \cos\frac{8\pi}{7}) +3(\cos\frac{6\pi}{7} + \cos\frac{12\pi}{7} +\cos\frac{24\pi}{7})$$
$$=(\cos\frac{2\pi}{7} +\cos\frac{4\pi}{7} + \cos(2\pi-\frac{8\pi}{7}) +3(\cos\frac{6\pi}{7} + \cos(2\pi-\frac{12\pi}{7}) +3\cos(4\pi-\frac{24\pi}{7})$$
$$=(\cos\frac{2\pi}{7} +\cos\frac{4\pi}{7} + \cos\frac{6\pi}{7}) +3(\cos\frac{6\pi}{7} + \cos\frac{2\pi}{7} +3\cos\frac{4\pi}{7})$$
$$=4(\cos\frac{2\pi}{7} +\cos\frac{4\pi}{7} + \cos\frac{6\pi}{7})$$
$$

\therefore \cos^3\frac{2\pi}{7} + \cos^3\frac{4\pi}{7} + \cos^3\frac{8\pi}{7}=\cos\frac{2\pi}{7} +\cos\frac{4\pi}{7} + \cos\frac{6\pi}{7}$$

If we let

$$x = \cos\frac{2\pi}{7} +\cos\frac{4\pi}{7} + \cos\frac{6\pi}{7}$$

Multiply through by $2\sin \dfrac{\pi}{7}$, we get

$$\begin{align*}2x\sin \frac{\pi}{7} &=2\cos\frac{2\pi}{7}\sin \frac{\pi}{7} + 2\cos\frac{4\pi}{7}\sin \frac{\pi}{7} + 2\cos\frac{6\pi}{7}\sin \frac{\pi}{7}\\&== sin \frac{3\pi}{7} - sin \frac{\pi}{7} + sin \frac{5\pi}{7} - sin \frac{3\pi}{7} + sin \frac{7\pi}{7} - sin \frac{5\pi}{7}\\&= sin \pi - sin \frac{\pi}{7}\\&=-\sin \frac{\pi}{7}\end{align*}$$$\therefore x = -\dfrac{1}{2}$

or

$$\cos^3\frac{2\pi}{7} + \cos^3\frac{4\pi}{7} + \cos^3\frac{8\pi}{7} = -\frac{1}{2}$$