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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:

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.

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This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

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## Comments

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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

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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.

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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

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

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