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)

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)

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TopNewestLet 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 · 3 years, 10 months ago

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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. – Calvin Lin Staff · 3 years, 10 months ago

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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 · 3 years, 10 months ago

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@Calvin: Thank you. I just didn't recognize that simple thing. Shame on me... – Đức Việt Lê · 3 years, 10 months ago

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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? – Calvin Lin Staff · 3 years, 10 months ago

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Thanks! – Pranav Arora · 3 years, 10 months ago

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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? – Calvin Lin Staff · 3 years, 10 months ago

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– Pranav Arora · 3 years, 10 months ago

Thank you but I am not a mathematics genius so I am still clueless here. I will wait for others to reply. :)Log in to reply

– Mursalin Habib · 3 years, 10 months ago

I'm sorry! I made a terrible mistake!. I don't know why (or how) I wrote that. So embarrassing!! Deleting comment.... And yes, I can show that....Log in to reply