Trigonometry #6: Find the Error

by djmathman, Nov 7, 2011, 2:08 AM

I have a proof for this problem, but I've been trying to find a second proof. It seems correct (and the answer is indeed correct), but I noticed an error somewhere in this proof. Can you find the error and help me correct it? Thanks. (I know where the error is; I just don't know how to fix it)
Problem wrote:
Find the maximum value of $\sin x + \cos x$ for $0\le x\le 2\pi$.

We look to the unit circle for inspiration. We see that the maximum will occur when $\sin x$ and $\cos x$ are both positive, so we focus on the first quadrant. For any right triangle in the unit circle, with leg lengths $a=\sin x$ and $b=\cos x$, we are given that $a^2+b^2=1$ since the radius of the circle is equal to $1$. Using $AM-GM$ on $a^2$ and $b^2$ we find that $\dfrac{a^2+b^2}{2}\ge \sqrt{a^2b^2}$, or $\dfrac{a^2+b^2}{2}\ge ab$. Since we know that $a^2+b^2=1$, we can substitute to get that $ab\le \dfrac{1}{2}$.

Now we use $AM-GM$ one last time, this time on $a$ and $b$. Doing so, we get $\dfrac{a+b}{2}\ge \sqrt{ab}$. Substituting our range for $ab$, we get $\dfrac{a+b}{2}\le \sqrt{\dfrac{1}{2}}$, or $\dfrac{a+b}{2}\le \dfrac{\sqrt{2}}{2}$. Finally, multiplying both sides of the inequality by $2$ gives us the desired $a+b\le\boxed{\sqrt{2}}$.

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On the last time you flipped the sign of AM-GM

To fix that, show that when you maximize $ab$, you maximize $a+b$

by sjaelee, Nov 7, 2011, 2:41 AM

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Okay...

OH WOW THIS FOLLOWS IN THE FOOTSTEPS OF THE OTHER PROOF I FOUND.

So we know that $(a+b)^2=a^2+2ab+b^2$. Since $a^2+b^2=1$, we know that $(a+b)^2=1+2ab$. Since by $AM-GM$ we have $ab\le \dfrac{1}{2}$, we have that $(a+b)^2\le 1+2\left(\dfrac{1}{2}\right)$, or $(a+b)^2\le 2$. Taking the positive square root of both sides gives us our answer.

(Note: the other proof I found was to let $\sin x + \cos x=k$, square, rearrange, and then bound.)

by djmathman, Nov 7, 2011, 2:47 AM

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  • dj so orz :omighty:

    by Yiyj1, Mar 29, 2025, 1:42 AM

  • legendary problem writer

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  • orz $$\,$$

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  • hi dj $ $ $ $

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  • i wanna submit my own problems lol

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  • hi dj, may i have the role of contributer? :D

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  • This was helpful!

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  • waiting for a recap of your amc proposals for this year :D

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  • also happy late bday man! i missed it by 2 days but hope you are enjoyed it

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  • :love: tfw kakuro appears on amc :love:

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  • Hi dj :)

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  • Roses are red,
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  • hello :)

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  • Do you have a link to your main blog that you started after graduating from high school, I couldn't find it. @dj I met you IRL at Awesome Math summer Program several years ago.

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