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Thread: differentiation of trigo functions

  1. #1
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    differentiation of trigo functions

    1.
    Given that y=2$\displaystyle sin^3$ x - 3 sin x, find an expression for $\displaystyle \frac{dy}{dx}$ in terms of cos x and then find $\displaystyle \frac{d^2y}{dx^2}$.

    2.
    The diagram shows an isosceles triangle ABC inscribed in a circle of radius r cm and centre at O. Given that $\displaystyle \angle {BAO}$=$\displaystyle \theta$ radian and $\displaystyle \angle$ ADC = $\displaystyle \frac{\pi}{2}$ radians, show that area , S $\displaystyle cm^2$, of triangle ABC is given by S = $\displaystyle r^2$ sin 2$\displaystyle \theta$(1+ cos 2$\displaystyle \theta$).


    for the first question, what i tried to do:
    $\displaystyle \frac{dy}{dx}$
    = 6($\displaystyle sin^2$ x)($\displaystyle cos^2$ x) - 3 cos x
    =3 cos x [ 2 $\displaystyle sin^2$ x cos x -1]

    then.....how do i continue to simplify it?the answer is 3 cos x(1- 2 $\displaystyle cos^2$ x)......


    for question 2 ....

    AB= $\displaystyle \frac{2r}{cos \theta}$


    S
    = ($\displaystyle \frac{1}{2}$)($\displaystyle \frac{2r}{cos \theta}$)($\displaystyle \frac{2r}{cos \theta}$)sin 2
    $\displaystyle \theta$
    =($\displaystyle \frac{2r^2}{cos^2 \theta}$) sin 2$\displaystyle \theta$

    and......then....how do i simplify it to $\displaystyle r^2$sin 2$\displaystyle \theta$(1+ cos 2$\displaystyle \theta$)....??? is my working correct in first place???



    any help is much appreciated. thanks in advance.
    Last edited by wintersoltice; Mar 9th 2009 at 06:14 AM.
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  2. #2
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    1. Redo your differentiation, cause it's false.
    Then there's a simple way of relating $\displaystyle sin^2(x)$ to $\displaystyle cos^2(x)$.

    2. The formula for AB is correct.
    Start from it and compute AD and BD.
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  3. #3
    Member u2_wa's Avatar
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    Quote Originally Posted by wintersoltice View Post
    1.
    Given that y=2$\displaystyle sin^3$ x - 3 sin x, find an expression for $\displaystyle \frac{dy}{dx}$ in terms of cos x and then find $\displaystyle \frac{d^2y}{dx^2}$.

    2.
    The diagram shows an isosceles triangle ABC inscribed in a circle of radius r cm and centre at O. Given that $\displaystyle \angle {BAO}$=$\displaystyle \theta$ radian and $\displaystyle \angle$ ADC = $\displaystyle \frac{\pi}{2}$ radians, show that area , S $\displaystyle cm^2$, of triangle ABC is given by S = $\displaystyle r^2$ sin 2$\displaystyle \theta$(1+ cos 2$\displaystyle \theta$).


    for the first question, what i tried to do:
    $\displaystyle \frac{dy}{dx}$
    = 6($\displaystyle sin^2$ x)($\displaystyle cos^2$ x) - 3 cos x
    =3 cos x [ 2 $\displaystyle sin^2$ x cos x -1]

    then.....how do i continue to simplify it?the answer is 3 cos x(1- 2 $\displaystyle cos^2$ x)......


    for question 2 ....

    AB= $\displaystyle \frac{2r}{cos \theta}$


    S
    = ($\displaystyle \frac{1}{2}$)($\displaystyle \frac{2r}{cos \theta}$)($\displaystyle \frac{2r}{cos \theta}$)sin 2
    $\displaystyle \theta$
    =($\displaystyle \frac{2r^2}{cos^2 \theta}$) sin 2$\displaystyle \theta$

    and......then....how do i simplify it to $\displaystyle r^2$sin 2$\displaystyle \theta$(1+ cos 2$\displaystyle \theta$)....??? is my working correct in first place???



    any help is much appreciated. thanks in advance.
    Let me solve both parts from the beginning:

    $\displaystyle y=2(sinx)^3-3sinx$
    $\displaystyle
    \frac{dy}{dx}=6(sinx)^2*cosx-3cosx$
    $\displaystyle =6(1-(cosx)^2)*cosx-3cosx$
    $\displaystyle =cosx((6(1-(cosx)^2)-3) $

    Now the second part:
    $\displaystyle \angle {BOD}=2\theta$ (it's circle property i.e angle twice at centre than at the corner)

    $\displaystyle AD(DO+OA)=rcos2\theta+r$
    $\displaystyle BC=2rsin2\theta$
    Area $\displaystyle = \frac{1}{2}(rcos2\theta+r)(2rsin2\theta)$
    $\displaystyle =r^2 sin 2\theta(1+ cos 2\theta)$

    Hope this helps
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  4. #4
    Member u2_wa's Avatar
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    To take $\displaystyle \frac{d^2y}{dx^2}$ take derivative of the $\displaystyle \frac{dy}{dx}$ function i.e calculated in first part of the previous post.
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