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# 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
4 years, 1 month 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 · 4 years, 1 month 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 · 4 years, 1 month 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 · 4 years, 1 month ago

@Calvin: Thank you. I just didn't recognize that simple thing. Shame on me... · 4 years, 1 month ago

Comment deleted May 29, 2013

If $$\alpha = \frac{\pi}{4}$$, what is $$\sin \alpha + \cos \alpha$$?

Your initially inequality is not true. It becomes true if you say that it is less than $$\sqrt{2}$$. Do you know how to show that? Staff · 4 years, 1 month ago

It is easy to show that $$\sin\alpha+\cos\alpha \leq \sqrt{2}$$ but how someone would prove the inequality asked by OP? Any hints for that?

Thanks! · 4 years, 1 month ago

There was a valid hint in the initial comment, which is that $$\sin \alpha \leq 1$$.

He wanted to use the (incorrect) fact that $$\sin \alpha + \cos \alpha \leq 1$$. Assuming this is true, then $$\sin^{2007} \alpha + \cos \alpha \leq \sin \alpha + \cos \alpha \leq 1 < \frac{5}{4}$$.
Of course, this is not true, and we instead get $$\sqrt{2}$$ on the RHS, which is not good enough.

Can you modify the argument slightly to make it work? Staff · 4 years, 1 month ago

Thank you but I am not a mathematics genius so I am still clueless here. I will wait for others to reply. :) · 4 years, 1 month ago