MHB Can You Solve the Cube of Cosines for Specific Angles?

[anemone]

Here is this week's POTW:

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Evaluate $$\cos^3 \left(\frac{2\pi}{7}\right)+\cos^3 \left(\frac{4\pi}{7}\right)+\cos^3 \left(\frac{8\pi}{7}\right).$$

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[anemone]

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

1. kaliprasad
2. MarkFL

Solution from MarkFL:

Spoiler

Let:

$$S=\cos^3\left(\frac{2\pi}{7}\right)+\cos^3\left(\frac{4\pi}{7}\right)+\cos^3\left(\frac{8\pi}{7}\right)$$

Now, if we use the power reductions formula:

$$\cos^3(A)=\frac{3\cos(A)+\cos(3A)}{4}$$

We may write:

$$4S=3\left(\cos\left(\frac{2\pi}{7}\right)+\cos\left(\frac{4\pi}{7}\right)+\cos\left(\frac{8\pi}{7}\right)\right)+\left(\cos\left(\frac{6\pi}{7}\right)+\cos\left(\frac{12\pi}{7}\right)+\cos\left(\frac{24\pi}{7}\right)\right)$$

Using:

$$\cos\left(\frac{6\pi}{7}\right)=\cos\left(\frac{8\pi}{7}\right)$$

$$\cos\left(\frac{12\pi}{7}\right)=\cos\left(\frac{2\pi}{7}\right)$$

$$\cos\left(\frac{24\pi}{7}\right)=\cos\left(\frac{4\pi}{7}\right)$$

We may now state:

$$S=\cos\left(\frac{2\pi}{7}\right)+\cos\left(\frac{4\pi}{7}\right)+\cos\left(\frac{8\pi}{7}\right)$$

Let:

$$r=\cos\left(\frac{2\pi}{7}\right)+i\sin\left(\frac{2\pi}{7}\right)$$

Now, consider the quadratic in $x$ with the following roots:

$$x_1=r+r^2+r^4$$

$$x_2=r^3+r^5+r^6$$

The sum of these roots is:

$$x_1+x_2=(1+r+r^2+r^3+r^4+r^5+r^6) - 1=\frac{1-r^7}{1-r}-1=-1$$

The product of the roots is:

$$x_1x_2=r^4+r^5+r^6+3r^7+r^8+r^9+r^10=1+r+r^2+r^3+r^4+r^5+r^6 + 2=2$$

Hence, our quadratic is:

$$x^2+x+2=0$$

And it's roots are:

$$x=\frac{-1\pm i\sqrt{7}}{2}$$

With:

$$x_1=r+r^2+r^4$$

We then see we must have:

$$S=-\frac{1}{2}$$