Basic Math Problem of the Week 11/09/2017

[PF PotW Robot]

Here is this week's basic math problem. We have several members who will check solutions, but we also welcome the community in general to step in. We also encourage finding different methods to the solution. If one has been found, see if there is another way. Using spoiler tags is optional. Occasionally there will be prizes for extraordinary or clever methods.

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

(PotW thanks to our friends at http://www.mathhelpboards.com/)

 

[jedishrfu]

I did a numerical check of the expression and got a value of -0.5 so that means there is some elegant way of simplifying it.

 

[fresh_42]

And the angles sum up to a full circle. I tried $4\cos^3 \varphi= 3\cos \varphi + \cos 3 \varphi$ to get rid of the powers, but I'm missing the clue. The sum becomes with this formula
$$
\cos^3 \dfrac{2}{7}\pi + \cos^3 \dfrac{4}{7}\pi + \cos^3 \dfrac{8}{7}\pi = \cos \dfrac{2}{7}\pi + \cos \dfrac{4}{7}\pi + \cos \dfrac{8}{7}\pi
$$
I assume the rest can be done by $\cos \varphi = \dfrac{1}{2} (\exp i\varphi + \exp -i\varphi)$.

 

[Charles Link]

And the angles sum up to a full circle. I tried $4\cos^3 \varphi= 3\cos \varphi + \cos 3 \varphi$ to get rid of the powers, but I'm missing the clue. The sum becomes with this formula
$$
\cos^3 \dfrac{2}{7}\pi + \cos^3 \dfrac{4}{7}\pi + \cos^3 \dfrac{8}{7}\pi = \cos \dfrac{2}{7}\pi + \cos \dfrac{4}{7}\pi + \cos \dfrac{8}{7}\pi
$$
I assume the rest can be done by $\cos \varphi = \dfrac{1}{2} (\exp i\varphi + \exp -i\varphi)$.

@fresh_42 That is very helpful. A google of cos(2 pi /7) showed a couple of identities with expressions like the one that you have that are solved by multiplying by $\frac{\sin(\pi/7)}{\sin(\pi/7)}$. See the video: $\\$ Additional comment:Our equation has an $8 \pi/7$ instead of the $6 \pi/7$ in the video. This makes the last terms in our case $\sin(9 \pi /7)-\sin(7 \pi/7)$ . We have $\sin(9 \pi /7)=-\sin(5 \pi/7)$, so the results are identical.

 

[jedishrfu]

That’s a nice video And the solution is quite elegant.

 

[StoneTemplePython]

Here is my approach:

let

$\lambda_1 = cos\big(\frac{2}{7}\pi\big)$
$\lambda_2 = cos\big(\frac{4}{7}\pi\big)$
$\lambda_3 = cos\big(\frac{8}{7}\pi\big)$

- - - -
key idea: symmetry

note $cos(a)$ obeys the identity: $cos^2(a) = \frac{1}{2}\big(1+ cos(2a)\big)$

so if we apply this to each $\lambda$, we get:
$\lambda_1^2 = \frac{1}{2}\big(1+ \lambda_2\big)$
$\lambda_2^2 = \frac{1}{2}\big(1+ \lambda_3\big)$
$\lambda_3^2 = \frac{1}{2}\big(1+ \lambda_1\big)$

Also noting that each $\lambda_k$ is the cosine of some multiple of pi, where each multiple is *not* a natural number, we know that $\big \vert \lambda_k \big \vert \lt 1$ for $k \in \{1,2,3\}$ -- i.e. each lambda has magnitude of less than one.

as mentioned by @fresh_42 we know that

$\lambda_1 + \lambda_2 + \lambda_3 = \lambda_1^3 + \lambda_2^3 + \lambda_3^3$

(which translates to the relation below)

$trace\big(\mathbf C\big) = trace\big(\mathbf C^3\big)$

- - - -
Solution:
From here I just beat this problem with a hammer using some facts from polynomial and matrix theory -- specifically things related to symmetric sums.

These $\lambda$'s are all real numbers -- let them be interpreted as roots to some polynomial in monic form, i.e.

$p(x) = x^3 + a_2 x^2 + a_1 x + a_0 = (x- \lambda_1)(x- \lambda_2)(x-\lambda_3)$

We can encode this polynomial in the companion matrix $\mathbf C$.

we know that

$a_2 = -trace\big(\mathbf C\big)$Newton's Identities (or just mechanical manipulation of the companion matrix) tell us that

$a_1 = -\frac{1}{2}\Big(a_2 trace\big(\mathbf C\big) + trace\big(\mathbf C^2\big)\Big) = -\frac{1}{2}\Big( -trace\big(\mathbf C\big)^2 + trace\big(\mathbf C^2\big)\Big) = \frac{1}{2} trace\big(\mathbf C\big)^2 - \frac{1}{2} trace\big(\mathbf C^2\big)$

but we also know that $a_1$ is a symmetric function given by

$a_1 = \lambda_1\lambda_2 + \lambda_2\lambda_3 + \lambda_1\lambda_3$

Now the symmetry exploit creeps in:

$trace\big(\mathbf C^3\big) = \lambda_1^3 + \lambda_2^3 + \lambda_3^3 = \lambda_1(\lambda_1^2) + \lambda_2(\lambda_2^2) + \lambda_3(\lambda_3^2) = \lambda_1\big(\frac{1}{2}( 1 + \lambda_2)\big) + \lambda_2\big(\frac{1}{2}( 1 + \lambda_3)\big) + \lambda_3\big(\frac{1}{2}( 1 + \lambda_1)\big)$$trace\big(\mathbf C\big) = trace\big(\mathbf C^3\big) = \frac{1}{2}\big(\lambda_1\lambda_2 + \lambda_2\lambda_3 + \lambda_3 \lambda_1\big) +\frac{1}{2}\big(\lambda_1 + \lambda_2+ \lambda_3 \big) = \frac{1}{2}\big(\lambda_1\lambda_2 + \lambda_2\lambda_3 + \lambda_3 \lambda_1\big) +\frac{1}{2}trace\big(\mathbf C \big)$

$\frac{1}{2} trace\big(\mathbf C\big) = \frac{1}{2} \big(\lambda_1\lambda_2 + \lambda_2\lambda_3 + \lambda_3 \lambda_1\big)$

$trace\big(\mathbf C\big) = a_1 = trace\big(\mathbf C^3\big)$

using the Newton identity, this gives us:

$a_1 = trace\big(\mathbf C\big) = \frac{1}{2} trace\big(\mathbf C\big)^2 - \frac{1}{2} trace\big(\mathbf C^2\big)$

- - - -

and we just need to make $trace\big(\mathbf C^2\big)$ 'go away'.

To finish this off, we examine the eigenvalue representation of $trace\big(\mathbf C^2\big)$ and again apply the symmetry exploit:

$trace\big(\mathbf C^2\big) = \lambda_1^2 + \lambda_2^2 + \lambda_3^2 = \frac{1}{2}\big(1 + \lambda_2\big) + \frac{1}{2}\big(1 + \lambda_3\big) + \frac{1}{2}\big(1 + \lambda_1\big) = \frac{3}{2} +\frac{1}{2}\big(trace\big(\mathbf C\big)\big)$

Now make the substitution into our equation from the Newton Identity.

$a_1 = trace\big(\mathbf C\big) = \frac{1}{2} trace\big(\mathbf C\big)^2 - \frac{1}{2} \Big(trace\big(\mathbf C^2\big)\Big) = \frac{1}{2} trace\big(\mathbf C\big)^2- \frac{1}{2}\Big( \frac{3}{2} +\frac{1}{2}trace\big(\mathbf C\big)\Big)$

multiply everything by 2

$2 a_1 = a_1^2- \Big( \frac{3}{2} +\frac{1}{2}a_1 \Big)$

$0 = a_1^2 - \frac{5}{2}a_1 - \frac{3}{2}$

And we can eyeball the trace and determinant of such an equation as having roots of $-\frac{1}{2}$ and $3$. Because each $\lambda$ has magnitude less than one, by triangle inequality, we know

$\big \vert \lambda_1 + \lambda_2 + \lambda_3 \big \vert\leq \big \vert \lambda_1\big \vert + \big \vert\lambda_2\big \vert + \big \vert\lambda_3\big \vert \lt 3$

and hence we throw out the root of 3.

We keep the root of

$-\frac{1}{2} = a_1 = trace\big(\mathbf C\big) =trace\big(\mathbf C^3\big) = \lambda_1^3 + \lambda_2^3 + \lambda_3^3$

as our answer, which completes the proof.