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Thread: maximum value of sine

  1. #1
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    maximum value of sine

    show that sinx (1+cosx) has a maximum value when x=Pie/3
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  2. #2
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    Quote Originally Posted by Shivanand View Post
    show that sinx (1+cosx) has a maximum value when x=Pie/3
    $\displaystyle f(x) = sin(x)(1+cos(x)) = sin(x) + sin(x)cos(x)$

    Differentiate!

    $\displaystyle f'(x) = cos(x)( \, 1+cos(x) \, ) + sin(x)(\, -sin(x) \, )$

    Maximum occurs when $\displaystyle f'(x) = 0$

    $\displaystyle f'(x) = cos(x)(1+cos(x) + sin(x)(-sin(x)) = 0$

    $\displaystyle cos(x) + cos^2(x) - sin^2(x) = 0$ (Remember that $\displaystyle cos^2(x) - sin^2(x) = cos(2x) $)

    $\displaystyle cos(x) + cos(2x) = 0$

    $\displaystyle cos(x) + 2cos^2(x) -1 = 0$ Remember that $\displaystyle cos(2x) =2cos^2(x) - 1$

    $\displaystyle cos(x)(1+2cos(x)) = 1$

    Take it from here?
    Last edited by mr fantastic; Dec 17th 2008 at 03:25 AM. Reason: Fixed the brackets
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  3. #3
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    Quote Originally Posted by Mush View Post
    $\displaystyle f(x) = sin(x)(1+cos(x)) {\color{red}= sin(x) + sin(x)cos(x)}$ Mr F says: The red stuff in this line is unnecessary (since you don't use it in your solution) and in fact unhelpful and potentially confusing to the OP.

    Differentiate!

    $\displaystyle f'(x) = cos(x)( \, 1+cos(x) \, ) + sin(x)(\, -sin(x) \, )$

    Maximum occurs when $\displaystyle f'(x) = 0$

    $\displaystyle f'(x) = cos(x)(1+cos(x) + sin(x)(-sin(x)) = 0$

    $\displaystyle cos(x) + cos^2(x) - sin^2(x) = 0$ (Remember that $\displaystyle cos^2(x) - sin^2(x) = cos(2x) $) Mr F says: Alternatively and more efficiently, remember the Pythagorean Identity $\displaystyle {\color{red}\cos^2 x + \sin^2 x = 1 \Rightarrow \sin^2 x = 1 - \cos^2 x}$. I'll pick up from here in my main reply below.

    $\displaystyle cos(x) + cos(2x) = 0$

    $\displaystyle cos(x) + 2cos^2(x) -1 = 0$ Remember that $\displaystyle cos(2x) =2cos^2(x) - 1$

    $\displaystyle cos(x)(1+2cos(x)) = 1$ Mr F says: This line does not lead anywhere helpful and may well reinforce a common misconception regarding the null factor law and its misuse.
    Take it from here?
    $\displaystyle \Rightarrow \cos x + 2 \cos^2 x -1 = 0$

    $\displaystyle \Rightarrow 2 \cos^2 x + \cos x - 1 = 0$

    $\displaystyle \Rightarrow (2 \cos x - 1)(\cos x + 1) = 0$

    and you should be able to see what happens now.
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  4. #4
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    Hello,

    $\displaystyle \sin(x)(1+\cos(x))=\sin(x)+\sin(x)\cos(x)=\sin(x)+ \frac 12 \sin(2x)$

    Differentiate :
    $\displaystyle \cos(x)+\cos(2x)$

    Set it equal to 0 :
    $\displaystyle \cos(x)+\cos(2x)=0$

    $\displaystyle \cos(2x)=\cos(\pi \pm x)$

    And we know that $\displaystyle \cos(a)=\cos(b) \Leftrightarrow a=b \text{ or } a=-b \quad +2k \pi$
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