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Geometry Level 3

Find the maximum value of the following expression. 32 cos ⁡ 6 x − 48 cos ⁡ 4 x + 18 cos ⁡ 2 x − 1 32\cos^6x-48\cos^4x+18\cos^2x-1 If the maximum value of the expression is equal to a a , enter the value of a 2 + 1 2 \dfrac{a^2+1}{2} .


The answer is 1.000.

Omkar Kulkarni by Omkar Kulkarni , 22, India

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1 solution

Omkar Kulkarni
Jun 24, 2015

This is simply the formula for cos ⁡ 6 x \cos6x in terms of cos ⁡ x \cos x , and hence the maximum value is 1 1 .

1 Helpful 0 Interesting 0 Brilliant 0 Confused

Moderator note:

And how are you suppose to show that?

In response to Challenge Master: Oh right. Completely forgot. :P

We use the double and tripe angle cosine formulas: cos ⁡ 3 x = 4 cos ⁡ 3 x − 3 cos ⁡ x \cos3x=4\cos^3x-3\cos x and cos ⁡ 2 x = 2 cos ⁡ 2 x − 1 \cos2x=2\cos^2x-1 .

cos ⁡ 6 x = cos ⁡ ( 3 ( 2 x ) ) = 4 cos ⁡ 3 ( 2 x ) − 3 cos ⁡ ( 2 x ) = 4 ( 2 cos ⁡ 2 x − 1 ) 3 − 3 ( 2 cos ⁡ 2 x − 1 ) = 4 ( 8 cos ⁡ 6 x − 1 − 12 cos ⁡ 4 x + 6 cos ⁡ 2 x ) − 6 c o s 2 x + 3 = 32 cos ⁡ 6 x − 4 − 48 cos ⁡ 4 x + 24 cos ⁡ 2 x − 6 cos ⁡ 2 x + 3 = 32 cos ⁡ 6 x − 48 cos ⁡ 4 x + 18 cos ⁡ 2 x − 1 \begin{aligned} \cos6x &=& \cos(3(2x)) \\ &=& 4\cos^3(2x)-3\cos(2x) \\ &=& 4(2\cos^2x-1)^3-3(2\cos^2x-1) \\ &=& 4(8\cos^6x-1-12\cos^4x+6\cos^2x)-6cos^2x+3 \\ &=& 32\cos^6x-4-48\cos^4x+24\cos^2x-6\cos^2x+3 \\ &=& 32\cos^6x-48\cos^4x+18\cos^2x-1 \end{aligned}

Omkar Kulkarni - 5 years, 11 months ago

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A nice problem and an elegant solution

Naman Kapoor - 5 years, 11 months ago

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Thank you! :D

Omkar Kulkarni - 5 years, 11 months ago

This is a Chebyshev polynomial of the 6th kind. Check out T6(x) in https://en.wikipedia.org/wiki/Chebyshev_polynomials

Vijay Simha - 1 year, 7 months ago

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