5
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Show that $$\left(1+4\sin^2\frac{\pi}{18}\right) \left(1+4\sin^2\frac{\pi}{6}\right) \left(1+4\sin^2\frac{5\pi}{18}\right)\left(1+4\sin^2\frac{7\pi}{18}\right)=34.$$

Any ideas about how to tackle this?

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    Do you know Viete's formula about roots of a polynomial. That's a place to start. – 2017-02-19
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    This annoys me because the pattern that sticks out to me is $x$, $3x$, $5x$, and $7x$ where $x=\frac\pi{18}$, and I've seen trig function identities like that before, but I can't look them up and I don’t remember how they work. – 2017-02-19
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    https://trans4mind.com/personal_development/mathematics/trigonometry/multipleAngles.htm – 2017-02-19
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    See http://math.stackexchange.com/questions/2145205/solution-trigonometry-of-equation – 2017-02-19
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    @labbhattacharjee How is that link useful? – 2017-02-19
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    @TakahiroWaki, The first link supplies the formula of $\cos9x$ and the second is a related problem but much tougher – 2017-02-19
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    Ok. I succeed, but too many riddles (sorry if I seem rude, I only want to be sarcastic :). We need at least two very (for me) long jumps to get here and, further, no much sense the one without "seeing" in advance the other. The first is how to pick the formula for $\cos9x$ among an infinity of them, the 18 in the denominator? The second is a variable change. Not considering other not-so-long jumps. It's said Viète or Vieta. – 2017-02-19

2 Answers 2

1

Got it!

Consider $T(x)=256X^4-576X^3+432X^2-120X+9$.

Then $\cos^2\frac{\pi}{18}$, $\cos^2\frac{\pi}{6}$, $\cos^2\frac{5\pi}{18}$, $\cos^2\frac{7\pi}{18}$ are its roots.

$34$ is then found using Viete's formulas.

$T$ is derived from the 9-th Chebyshev polynomial.

1

The 9th Chebyshev polynomial:

$$\cos(9x)=256\cos^9x-576\cos^7x+432\cos^5x-120\cos^3x+9\cos x$$

$$\frac{\cos(9x)}{\cos x}=256\cos^8x-576\cos^6x+432\cos^4x-120\cos^2x+9$$

$$\frac{\cos(9x)}{\cos x}=256(\cos^2x)^4-576(\cos^2x)^3+432\cos^4x-120\cos^2x+9$$

$$256(\cos^2x)^4-576(\cos^2x)^3+432(\cos^2x)^2-120\cos^2x+9=0\implies \cos(9x)=0$$

Then:

$$9x=\pi/2,3\pi/2,5\pi/2,7\pi/2\;;x=\pi/18,3\pi/18,5\pi/18,7\pi/18$$

So, at least $\cos^2\frac{\pi}{18},\cos^2\frac{3\pi}{18},\cos^2\frac{3\pi}{18},\cos^2\frac{7\pi}{18}$ are solutions of:

$$256X^4-576X^3+432X^2-120X+9=0\tag 1$$

Now, we have $1+4\sin^2x=1+4(1-\cos^2x)=5-4\cos^2x$ And let be $X=\cos^2x$ and $t=1+4\sin^2x$ then $X=(5-t)/4$

$$256\left(\frac{5-t}{4}\right)^4-576\left(\frac{5-t}{4}\right)^3+432\left(\frac{5-t}{4}\right)^2-120\left(\frac{5-t}{4}\right)+9=0$$

Or

$$t^4-11t^3+42t^2-65t+34=0\tag 2$$

For each $X$ solution of (1) we have one $t$ solution of (2). e.g $X=\cos^2(\pi/18)$ solution of (1) makes $t=1+4\sin^2(\pi/18)$ solution of (2)

Now, let be $t_1,t_2,t_3,t_4$ the four different solutions of (2), then $t_1t_2t_3t_4=34$, or

$$\left(1+4\sin^2\frac{\pi}{18}\right) \left(1+4\sin^2\frac{\pi}{6}\right) \left(1+4\sin^2\frac{5\pi}{18}\right)\left(1+4\sin^2\frac{7\pi}{18}\right)=34$$