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Math Help - Demonstration involving sine and cosine

  1. #1
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    Demonstration involving sine and cosine

    Hi there, I want to demonstrate that

    1\leq{|\cos x|+|\sin x|\leq{\sqrt[]{2}}}

    How should I proceed?

    Bye and thanks.
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  2. #2
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    Quote Originally Posted by Ulysses View Post
    Hi there, I want to demonstrate that

    1\leq{|\cos x|+|\sin x|\leq{\sqrt[]{2}}}

    How should I proceed?

    Bye and thanks.
    You can use a unit circle and/or the identity \cos^2x+\sin^2x=1.
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    How do I fit that trig identity intro my inequality?

    Thanks.
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  4. #4
    MHF Contributor undefined's Avatar
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    Quote Originally Posted by Ulysses View Post
    How do I fit that trig identity intro my inequality?

    Thanks.
    Hmm I wonder if I'm overcomplicating things.

      |\cos x| + |\sin x| = |\cos x| + \sqrt{1-\cos^2x} = |\cos x| + \sqrt{1-|\cos x|^2}

    Recall also that 0\le |\cos x|\le 1

    You can substitute y = |\cos x| and show that the maximum on this interval is approximately \sqrt{2} using a graphing calculator (I'm assuming you haven't learned calculus yet, which you could use to find the maximum exactly). Since it says "demonstrate" and not "prove" this should be enough.
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  5. #5
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    I've learn some calculus, I'm actually trying to use this for a limit involving two variables :P I thought of using max and min, but I didn't cause I thought it would appear the sine and the cosine again driving me nowhere.
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  6. #6
    MHF Contributor Unknown008's Avatar
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    For the minimum, it's easy to find it. Put x = 0, then cos(x) = 1, and sin(x) = 0. 1 + 0 = 1.

    For the maximum, you might not be familiar with what value of x to make cos(x) + sin(x) maximum.

    I would find the derivative of the function first.

    \dfrac{d(cos(x)+sin(x))}{dx} = -sin(x) + cos(x)= cos(x) - sin(x)

    Then, solve for zero to find the maximum and minimum values.

    However, we'll have to convert this into a simpler form.

    cos(x) - sin(x) = Rcos(x + \alpha)

    Where R and alpha are constants.

    Expanding on the right side gives:

    R[cos(x)cos(\alpha) - sin(x)sin(\alpha)] = Rcos(x)cos(\alpha) - Rsin(x)sin(\alpha)

    Equate:

    cos(x) = Rcos(x)cos(\alpha)
    -sin(x) = - Rsin(x)sin(\alpha)

    Giving:

    1= Rcos(\alpha) ---1
    1 = Rsin(\alpha) ---2

    Take 2/1:
    1 = tan(\alpha)

    \alpha = 45^o

    Take 1^2 + 2^2;

    1^2 + 1^2 = R^2cos^2(\alpha) + R^2sin^2(\alpha)

    2 = R^2

    R = \sqrt{2}

    So, cos(x) - sin(x) = \sqrt{2}cos(x + 45^o)

    Set to zero now:

    \sqrt{2}cos(x + 45^o) = 0

    cos(x + 45^o) = 0

    The maximum value of this solution is when x + 45^o = 90^o

    x = 45^o
    Last edited by Unknown008; September 13th 2010 at 09:12 AM. Reason: For some reason, when I clicked on preview, your previous posts didn't appear... =/ Sorry for that.
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