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Math Help - Trigonometric Differentiation

  1. #1
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    Trigonometric Differentiation

    Given that x = 4sin(2y+6), find (dy/dx) in terms of x.

    i have managed to get:

    (dy/dx) = 1/8cos(2y+6)

    where do i go from here?
    Last edited by Stylis10; June 5th 2008 at 07:02 AM.
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  2. #2
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    Quote Originally Posted by Stylis10 View Post
    Given that x = 4sin(2y+6), find (dy/dx) in terms of x.

    i have managed to get:

    (dy/dx) = 1/8cos(2y+6)

    where do i go from here?
    \cos(2y+6) = \pm \sqrt{1 - \sin^2 (2y+6)} = \pm \sqrt{1 - \left(\frac{x}4\right)^2}

    I dont know what to do with the sign of cos though. It can be both plus and minus...
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  3. #3
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    Quote Originally Posted by Isomorphism View Post
    \cos(2y+6) = \pm \sqrt{1 - \sin^2 (2y+6)} = \pm \sqrt{1 - \left(\frac{x}4\right)^2}

    I dont know what to do with the sign of cos though. It can be both plus and minus...
    ok i under stand the first step of that, but can u please explain to me how you managed to get (x/4)^2 from sin^2(2y+6)
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  4. #4
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    ok i understand how you got that, thanxs
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  5. #5
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    Hello, Stylis10!

    Given: . x \:= \:4\sin(2y+6), .find \frac{dy}{dx} in terms of x.

    i have managed to get: . \frac{dy}{dx} \:= \:\frac{1}{8\cos(2y+6)} . . . . Good!

    where do i go from here?

    We are given: . 4\sin(2y+6) \:=\:x\quad\Rightarrow\quad \sin(2y+6) \:=\:\frac{x}{4} \:=\:\frac{opp}{hyp}

    So (2y+6) is an angle in a right triangle with: . opp = x,\;hyp = 4
    . . From Pythagorus, we have: . adj = \sqrt{16-x^2}
    Hence: . \cos(2y+6) \:=\:\frac{\sqrt{16-x^2}}{4}

    Substitute into your equation: . \frac{dy}{dx} \;=\;\frac{1}{8\!\cdot\!\frac{\sqrt{16-x^2}}{4}}

    Therefore: . \boxed{\frac{dy}{dx}\;=\;\frac{1}{2\sqrt{16-x^2}}}


    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

    For the fun of it, I did it another way.
    . . (I've got to get a life!)

    We have: . 4\sin(2y+6) \;=\;x \quad\Rightarrow\quad \sin(2y+6) \:=\:\frac{x}{4}

    . . . 2y + 6 \;=\;\sin^{-1}\left(\frac{x}{4}\right) \quad\Rightarrow\quad y \;=\;\frac{1}{2}\sin^{-1}\left(\frac{x}{4}\right) - 3


    Then: . \frac{dy}{dx} \;\;=\;\;\frac{1}{2}\cdot\frac{\frac{1}{4}}{\sqrt{  1-\left(\frac{x}{4}\right)^2}} \;\;=\;\; \frac{1}{8}\!\cdot\!\frac{1}{\sqrt{1-\frac{x^2}{16}}}

    . . . . . \frac{dy}{dx} \;\;=\;\;\frac{1}{8}\!\cdot\!\frac{1}{\sqrt{\frac{  16-x^2}{16}}} \;\;=\;\;\frac{1}{8}\!\cdot\!\frac{1}{\frac{\sqrt{  16-x^2}}{4}}


    Therefore: . \frac{dy}{dx} \;=\;\frac{1}{2\sqrt{16-x^2}}\quad\hdots  ta-DAA!

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