An inequality

Today I met a prob:

"Prove that $(sin \alpha)^{2007} + cos \alpha < \frac{5}{4}$ (With $\alpha$ 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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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

- 6 years, 4 months ago

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

All that you need to show is that $\sin^2 \alpha + \cos \alpha = 1 + \cos \alpha - \cos^2 \alpha \leq \frac{5}{4}$, 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.

Staff - 6 years, 4 months ago

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

- 6 years, 4 months ago

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

- 6 years, 4 months ago