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Math Help - differentiation of trigo functions

  1. #1
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    differentiation of trigo functions

    1.
    Given that y=2 sin^3 x - 3 sin x, find an expression for \frac{dy}{dx} in terms of cos x and then find \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 \angle {BAO}= \theta radian and \angle ADC = \frac{\pi}{2} radians, show that area , S cm^2, of triangle ABC is given by S = r^2 sin 2 \theta(1+ cos 2 \theta).


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

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


    for question 2 ....

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


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

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



    any help is much appreciated. thanks in advance.
    Last edited by wintersoltice; March 9th 2009 at 07:14 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 sin^2(x) to 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 sin^3 x - 3 sin x, find an expression for \frac{dy}{dx} in terms of cos x and then find \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 \angle {BAO}= \theta radian and \angle ADC = \frac{\pi}{2} radians, show that area , S cm^2, of triangle ABC is given by S = r^2 sin 2 \theta(1+ cos 2 \theta).


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

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


    for question 2 ....

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


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

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



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

    y=2(sinx)^3-3sinx
    <br />
\frac{dy}{dx}=6(sinx)^2*cosx-3cosx
    =6(1-(cosx)^2)*cosx-3cosx
    =cosx((6(1-(cosx)^2)-3)

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

    AD(DO+OA)=rcos2\theta+r
    BC=2rsin2\theta
    Area = \frac{1}{2}(rcos2\theta+r)(2rsin2\theta)
     =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 \frac{d^2y}{dx^2} take derivative of the \frac{dy}{dx} function i.e calculated in first part of the previous post.
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