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Math Help - [SOLVED] Determing absolute minimum and maximum values

  1. #1
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    [SOLVED] Determing absolute minimum and maximum values

    f(x)=sin(x)+cos(x) on the integral [0, \frac{\pi}{3}]
    f'(x) = cos(x) - sin(x) = 0
    \frac{sin(x)}{cos(x)}=tan(x)=1=\frac{\pi}{4}

    0, (\frac{\pi}{4}), (\frac{\pi}{3})

    f(0)=sin(1)+cos(1)=0+1=1
    But how do I find out these two calculations below without using a calculator?
    f(\frac{\pi}{4}) = sin(\frac{\pi}{4}) + cos(\frac{\pi}{4})=\sqrt{2} = 1.44
    and
    f(\frac{\pi}{3}) = sin(\frac{\pi}{3}) + cos(\frac{\pi}{3}) = \sqrt{3} + 1 / 2 = 1.37
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    Quote Originally Posted by !!! View Post
    f(x)=sin(x)+cos(x) on the integral [0, \frac{\pi}{3}]
    f'(x) = cos(x) - sin(x) = 0
    \frac{sin(x)}{cos(x)}=tan(x)=1=\frac{\pi}{4}

    0, (\frac{\pi}{4}), (\frac{\pi}{3})

    f(0)=sin(1)+cos(1)=0+1=1
    But how do I find out these two calculations below without using a calculator?
    f(\frac{\pi}{4}) = sin(\frac{\pi}{4}) + cos(\frac{\pi}{4})=\sqrt{2} = 1.44
    and
    f(\frac{\pi}{3}) = sin(\frac{\pi}{3}) + cos(\frac{\pi}{3}) = \sqrt{3} + 1 / 2 = 1.37
    The points you have to check to know which are largest and smallest are: 1,\sqrt{2}, (\sqrt{3}+1)/2. To determine which is largest perform the same operations on this triple. First multiple each by 2: 2,2\sqrt{2}, \sqrt{3}+1. Now square each one: 4,8,4+\sqrt{3}. Subtract 4 from each one: 0,4,\sqrt{3}. And square again: 0,16,3. We see that that 1st point is the smallest, 3rd point is in between, and 2nd point is largest.
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  3. #3
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    Ahh, thank you.
    But how do I determine \sqrt{2} from sin(\frac{\pi}{4}) + cos(\frac{\pi}{4}) and \sqrt{3} + 1 / 2 from sin(\frac{\pi}{3}) + cos(\frac{\pi}{3})?
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    Quote Originally Posted by !!! View Post
    Ahh, thank you.
    But how do I determine \sqrt{2} from sin(\frac{\pi}{4}) + cos(\frac{\pi}{4}) and \sqrt{3} + 1 / 2 from sin(\frac{\pi}{3}) + cos(\frac{\pi}{3})?
    Because (you are expected to know) \sin \frac{\pi}{4} = \cos \frac{\pi}{4} = \frac{\sqrt{2}}2. And \cos \frac{\pi}{3} = \frac12 and \sin \frac{\pi}{3} = \frac{\sqrt{3}}2
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