An inequality

Today I met a prob:

"Prove that (sinα)2007+cosα<54 (sin \alpha)^{2007} + cos \alpha < \frac{5}{4} (With α \alpha is an acute angle)

Please tell me the way to solve the prob. (or if the prob's wrong)

Note by Đức Việt Lê
6 years, 4 months ago

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5 votes

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Let cos a = sqrt (1- sin^2 a)

We now want to find the maximum of x^2007 + sqrt(1-x^2) where x is in the range (0,1)

Using some calculus right away, you can find the maximum, but that would be overkill...

x^2007 + sqrt(1-x^2) < x^2 + sqrt(1-x^2)

Replace x^2 with y

we want to maximise y + sqrt (1-y). This is far easier to attack with calculus, and has a global maximum of 5/4, when y = 3/4. This gives us the desired result, or we could complete the square:

Replace 1-y with z^2:

y + sqrt (1-y) - 5/4 = (1-z^2) + z - 5/4 = -z^2 + z - 1/4 = (-1/4) ( 4z^2 - 4z + 1) = (-1/4) (2z-1)^2 <= 0,

this gives us the desired result

Gabriel Wong - 6 years, 4 months ago

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You're missing the mark, while coming close to it in numerous ways. Remember that sin2α\sin^2 \alpha can be replaced with 1cos2α 1-\cos^2 \alpha.

All that you need to show is that sin2α+cosα=1+cosαcos2α54 \sin^2 \alpha + \cos \alpha = 1 + \cos \alpha - \cos^2 \alpha \leq \frac{5}{4} , which follows directly from completing the square.

This in essence is what you did, though through a convoluted substitution that hides the simplicity of it.

Calvin Lin Staff - 6 years, 4 months ago

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Ah yes, you are right. By substituting on the sine initially, it made things slightly harder.

After finishing though, it is easy (if tidiness is desired) to go back and express z as the cosine, which gives us the (simpler) one line solution:

sin^2007(x) + cosx < sin^2(x) + cosx = 1 + cosx - cos^2(x) <= 5/4

Gabriel Wong - 6 years, 4 months ago

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@Calvin: Thank you. I just didn't recognize that simple thing. Shame on me...

Đức Việt Lê - 6 years, 4 months ago

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