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    Hi, I can't find a solution bank for the Edecel FP2 exam style paper so though i'd ask here.

    Here's the question: Name:  maths 1.png
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    I set x = rcos(theta)

    then plug in r and differentiate, setting the differential equal to 0. But I cant get the expression in terms of just sin or cos.

    using various Trig identities I've ended up with:

    1/2 = (sin(theta))^2 + (sq/(3))(sin(theta)cos(theta))

    help!
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    (Original post by Sir_Malc)
    Hi, I can't find a solution bank for the Edecel FP2 exam style paper so though i'd ask here.

    Here's the question: Name:  maths 1.png
Views: 70
Size:  395.1 KB

    I set x = rcos(theta)

    then plug in r and differentiate, setting the differential equal to 0. But I cant get the expression in terms of just sin or cos.

    using various Trig identities I've ended up with:

    1/2 = (sin(theta))^2 + (sq/(3))(sin(theta)cos(theta))

    help!
    \displaystyle x=r\cos \theta = \sin \theta \cos \theta + \sqrt{3}\cos^2 \theta = \frac{1}{2}\sin 2\theta + \sqrt{3}\cos^2 \theta.

    Differentiate:

    \displaystyle \cos 2\theta - 2\sqrt{3}\sin \theta \cos \theta = 0 \iff \cos 2\theta - \sqrt{3}\sin 2\theta = 0. Then re-arrange to get tan 2theta = something.

    There are other ways to do this, of course, but this is the second thing that popped into my head.
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    Ahh, I didn't think about changing them both in to 2theta, and that allows you to divide by cos2theta and make tan2theta.

    Thankyou very much!
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    (Original post by Sir_Malc)
    Ahh, I didn't think about changing them both in to 2theta, and that allows you to divide by cos2theta and make tan2theta.

    Thankyou very much!
    No problem.
 
 
 
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