A Trig Identity

Question

There are many trig identities that can be proven with known identities, but often the identities are tricky in finding the right methods to use. This is one of those identities that seems easy, but elusive. Consider the cubic equation $$ x^3 - 14 x^2 + 56 x - 56 = 0$$ for which one solution is $8 \sin^2(2\pi/7)$ and is equal to $4.89008...$. Now, another solution can be found to be $$7 - 6 \cos(\pi/7) + 6 \cos(2\pi/7) - 2 \cos(3\pi/7)$$ which is also found to have the numerical value $4.89008...$. Since both forms have the same numerical value and satisfy the same equation they must be equal. This leads to the identity $$8 \sin^2(2\pi/7) = 7 - 6 \cos(\pi/7) + 6 \cos(2\pi/7) - 2 \cos(3\pi/7).$$

The purpose of this question is to ask what methods are best suited to prove this equation by use of trig identities, or a proof of this identity.

Use the Pythagorean identity followed by reducing the remaining to be powers of 1 and expanding everything? — Simply Beautiful Art, Aug 20 '16 at 23:18
Maybe if you can, it may simply as easy as finding the closed form of each in terms of radicals and all that. — Simply Beautiful Art, Aug 20 '16 at 23:23
Double and triple angle formulas would probably also come in handy. — Bobson Dugnutt, Aug 20 '16 at 23:25

score 1 · Answer 1 · answered Aug 21 '16 at 02:01

1

Essentially you want to prove a linear relation between $\cos(k\pi/7)$ for $k=0,1,2,3$.

All those relations are multiples of the one that says that the average $x$-coordinate of the vertices of an inscribed regular $7$-gon is $0$.

answered Aug 21 '16 at 02:01

zyx

35,436

score 1 · Answer 2 · edited Apr 13 '17 at 12:21

Writing $7x=\pi$

We need to prove $$7-6\cos x+6\cos2x-2\cos3x=8\sin^22x=4(1-\cos4x)\ \ \ \ (1)$$

Now as $\cos2x=\cos(\pi-5x)=-\cos5x,\cos4x=\cdots=-\cos3x$

$(1)$ reduces to $$\cos x+\cos3x+\cos5x=\dfrac12$$

Can you use Prove that $\cos\frac {2\pi}{7}+ \cos\frac {4\pi}{7}+ \cos\frac {8\pi}{7}=-\frac{1}{2}$ to prove this?

Now using Trigonometry Sine Multiple Angle Formulae,

$$\sin7x=7\sin x-56\sin^3x+112\sin^5x-64\sin^7x$$

Now $\sin7x=0,7x=m\pi$ where $m$ is any integer, $x=\dfrac{m\pi}7$ where $m\equiv0,\pm1,\pm2,\pm3\pmod7$

So, the roots of $7s-56s^3+112s^5-64s^7=0$ are $\sin\dfrac{m\pi}7$ where $m\equiv0,\pm1,\pm2,\pm3\pmod7$

As $\sin0=0,$ the roots of $7-56(s^2)+112(s^2)^2-64(s^2)^3=0\ \ \ \ (1)$ are $\sin\dfrac{m\pi}7$ where $m\equiv\pm1,\pm2,\pm3\pmod7$

Let $y=8\sin^2\dfrac{2\pi}7,$ as $\sin\dfrac{2\pi}7$ is a root of $(1)$

$$7-56\cdot\dfrac y8+112\left(\dfrac y8\right)^2-64\left(\dfrac y8\right)^3=0$$

Can you take it from here?

A Trig Identity

2 Answers2