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Math Help - Differentiation of trigonometry equations

  1. #1
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    Differentiation of trigonometry equations

    Given that y=\cos2x + sin2x, find \frac{d^2y}{dx^2} and show that \frac{d^2y}{dx^2} + 4y = 0

    Please help. I have tried to use the two trig calculus rules that I know but I can't seem to get them to work.
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  2. #2
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    We know that \frac{{d^2 y}}{{dx^2 }} =  - 4\cos (2x) - 4\sin (2x).
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  3. #3
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    Hello, Greg!

    You must be differentiating incorrectly . . .


    Given that: . y\:=\:\cos2x + \sin2x, .find \frac{d^2y}{dx^2}

    and show that: . \frac{d^2y}{dx^2} + 4y \:=\: 0

    We have: . \begin{array}{ccc}y &=& \cos2x + \sin2x \\ \\ [-3mm]<br />
\dfrac{dy}{dx} &=& \text{-}2\sin2x + 2\cos2x \\ \\[-3mm]<br />
\dfrac{d^2y}{dx^2} &=& \text{-}4\cos2x - 4\sin2x \end{array}


    Therefore: . \begin{array}{ccccccc}<br />
\dfrac{d^2y}{dx^2} &=& \text{-}4\cos2x - 4\sin 2x \\ \\[-3mm]<br />
+4y\;\; &=& 4\cos2x + 4\sin2x \\ \hline \\[-4mm]<br /> <br />
\dfrac{d^2y}{dx^2} + 4y &=& 0\qquad\qquad\qquad <br />
\end{array}

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  4. #4
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    Can you tell me how you got from cos2x to -2sin2x?
    I know that cosx = -sinx but the 2x confused me
    Last edited by greghunter; June 27th 2009 at 11:39 AM.
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  5. #5
    MHF Contributor Amer's Avatar
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    \frac{d}{dx}(cos(f(x)))=f'(x)(-sin(f(x)) chain rule

    \frac{d}{dx}(cos(x^2))=2x^2(-sin(x^2))

    \frac{d}{dx}(cos(5x-1))=\frac{d}{dx}(5x-1) (-sin(5x-1))=5 (-sin(5x-1))
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  6. #6
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    Quote Originally Posted by Amer View Post
    \frac{d}{dx}(cos(f(x)))=f'(x)(-sin(f(x)) chain rule

    \frac{d}{dx}(cos(x^2))=2x^2(-sin(x^2))

    \frac{d}{dx}(cos(5x-1))=\frac{d}{dx}(5x-1) (-sin(5x-1))=5 (-sin(5x-1))
    thanks, i don't think i've been taught the chain rule yet
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