Chebyshev Polynomial Practice: Product

Algebra Level 3

Evaluate

$\cos(10^{\circ})\cos(20^{\circ})\cos(30^{\circ})\cos(40^{\circ})\cos(50^{\circ})\cos(60^{\circ})\cos(70^{\circ})\cos(80^{\circ})$

If the answer can be expressed as $\dfrac{a}{b}$

Find $a+b$

The answer is 259.

2 solutions

Brian Charlesworth
Apr 13, 2015

Since $\cos(90 - x) = \sin(x),$ this product can be written as

$\sin(10)\cos(10)\sin(20)\cos(20)\sin(30)\cos(30)\sin(40)\cos(40)$

$= \dfrac{1}{16}\sin(20)\sin(40)\sin(60)\sin(80),$

where the identity $\sin(2x) = 2\sin(x)\cos(x)$ was used. We also have the identity

$\sin(x)\sin(60 - x)\sin(60 + x) = \dfrac{1}{4}\sin(3x).$

With $x = 20$ our expression then becomes

$\dfrac{1}{16}*\dfrac{1}{4}*\sin(3*20)\sin(60) = \dfrac{1}{64}\left(\dfrac{\sqrt{3}}{2}\right)^{2} = \dfrac{3}{256},$

and thus $a + b = 3 + 256 = \boxed{259}.$

@Trevor Arashiro I figured you'll be writing up the Chebyshev solution method some time so I thought I'd just post this alternative approach for sake of variety. Are you learning about Chebyshev polynomials in school or on your own initiative?

P.S.. Did you watch The Masters? Spieth was pretty impressive, huh.

Brian Charlesworth - 6 years, 2 months ago

I'm learning them out side of school. Luckily I have had a lot of help on this topic. There will be a new skill coming out for these polynomials so I figured I'd go ahead and post a few questions to add to that section.

Speith was amazing. All sorts of records are being broken this year and last. Personally, my favorite golfer is Bubba Watson, who's yours?

Trevor Arashiro - 6 years, 2 months ago

There's such a good crop of young golfers right now it's hard to choose a favorite. I admire talent and ambition when it's tempered by genuine humility, so of the young golfers McIlroy, Spieth and Fowler would top my list, with Spieth probably at the top. I would love to see Mickelson finally win the U.S. Open this year; I didn't like him much early in his career but he has really grown on me. If I were starting out and looking to copy a swing I would choose Adam Scott's, (except for his putting stroke). He also seems like an exceptionally nice person. Bubba is a hoot and he wears his heart on his sleeve, but he can be a bit abrasive. He can pull off shots no one else can, though.

My all-time fave is Jack Nicklaus, of course, closely followed by Ben Hogan. I learned golf through their books, and for different reasons found them both to be great role models. :)

Brian Charlesworth - 6 years, 2 months ago

Trevor Arashiro
Apr 13, 2015

Let $T_n$ be the nth Chebyshev Polynomial where $T_n(\cos(x))=\cos(nx)$

Break up the equation into 3 parts.

$(\cos(10)\cos(50)\cos(70))\cdot(\cos(20)\cos(40)\cos(80))\cdot(\cos(30)\cos(60))$

Focusing on the first set, we have

$\cos(10)\cos(110)\cos(130)$ . Now we are looking at

$T_3(\cos(x))-\frac{\sqrt3}{2}=4x^3-3x-\frac{\sqrt3}{2}=0$

Vietas yields $\dfrac{\sqrt3}{8}$

Focusing on the second set

$\cos(20)\cos(100)\cos(140)$ . We have

$T_3(\cos(x))-\frac{1}{2}=4x^3-3x-\frac{1}{2}=0$

Vietas yields $\dfrac{1}{8}$

The third set can be calculated by hand and is equal to $\frac{\sqrt3}{4}$

Multiplying all these together we get

$\dfrac{\sqrt3}{8}\cdot\dfrac{1}{8}\cdot\dfrac{\sqrt3}{4}=\dfrac{3}{256}$

Chebyshev Polynomial Practice: Product

The answer is 259.

2 solutions

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