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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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    Re: [SOLVED] Determing absolute minimum and maximum values

    Quote Originally Posted by ThePerfectHacker View Post
    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
    Construct a right triangle with legs each of length 1. Since this is an isosceles triangle, the two acute angles are congruent. Since they must add to \pi/2, they are each \pi/4. The hypotenuse has, by the Pythagorean theorem, length \sqrt{2} so that sin(\pi/4) equals "opposite side over hypotenuse" which is \frac{1}{\sqrt{2}}= \frac{\sqrt{2}}{2} and the same for the cosine.

    For \pi/6 and \pi/3 construct an equilateral triangle (all three side of equal length so all three angles congruent so each angle \frac{\pi}{3}) with sides of length 2 and draw a line from one vertex perpendicular to the opposites side. By symmetry (that line divides the equilateral triangle into two congruent right triangles) we have a right triangle with one angle (\pi/3)/2= \pi/6, another angle \pi/3, hypotenuse of length 2 and side opposite the [tex]\pi/6[itex] angle of length 1. By the Pythagorean theorem the other leg (opposite the \pi/3 angle and the perpendicular constructed) has length \sqrt{4- 1}= \sqrt{3}.

    From that, we have sin(\pi/6)= \frac{1}{2}, cos(\pi/6)= \frac{\sqrt{3}}{2} and sin(\pi/3)= \frac{\sqrt{3}}{2}, cos(\pi/3)= \frac{1}{2}.
    Last edited by HallsofIvy; August 7th 2014 at 06:06 AM.
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