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Thread: solving trigo equation

  1. #1
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    solving trigo equation

    $\displaystyle 5cos^2 \theta + 12 cos \theta sin \theta = a + b cos (2\theta + y)$
    Find a , b and y.


    what i tried to do is:
    $\displaystyle 5cos^2 \theta + 12 cos \theta sin \theta$
    $\displaystyle =5(cos 2\theta + sin^2 \theta)+ 6(2 cos \theta sin \theta) $
    $\displaystyle =5sin^2 \theta + 5 cos 2\theta + 6sin2\theta$
    $\displaystyle =5sin^2 \theta + \sqrt{61} cos (2\theta-50.2^\circ)$


    then....stuck....is $\displaystyle a= 5sin^2 \theta$ ?
    or it can be further simplified/manipulated??
    $\displaystyle a= 5sin^2 \theta$ doesn't seem to be the final answer.....
    Last edited by wintersoltice; Aug 28th 2009 at 10:46 PM.
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  2. #2
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    The question asks you to write $\displaystyle 5\cos^2\theta+12\cos\theta\sin\theta$ in terms of $\displaystyle 2\theta$. So I would start by using the formulas $\displaystyle \cos^2\theta = \tfrac12(1+\cos(2\theta))$ and $\displaystyle 2\cos\theta\sin\theta = \sin(2\theta)$. That gives

    $\displaystyle 5\cos^2\theta+12\cos\theta\sin\theta = \tfrac52(1+\cos(2\theta)) + 6\sin(2\theta)$. Next step would be to write this as $\displaystyle \tfrac52 + \tfrac{13}2\bigl(\tfrac5{13}\cos(2\theta) + \tfrac{12}{13}\sin(2\theta)\bigr).$ Can you see why, and how to take it from there?
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  3. #3
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    Quote Originally Posted by wintersoltice View Post
    $\displaystyle 5cos^2 \theta + 12 cos \theta sin \theta = a + b cos (2\theta + y)$
    Find a and b.
    Hence, solve $\displaystyle 5cos^2 \theta + 12 cos \theta sin \theta = 0$.

    what i tried to do is:
    $\displaystyle 5cos^2 \theta + 12 cos \theta sin \theta$
    $\displaystyle =5(cos 2\theta + sin^2 \theta)+ 6(2 cos \theta sin \theta) $
    $\displaystyle =5sin^2 \theta + 5 cos 2\theta + 6sin2\theta$
    $\displaystyle =5sin^2 \theta + \sqrt{61} cos (2\theta-50.2^\circ)$
    Hi

    $\displaystyle 5 \cos^2 \theta + 12 \cos \theta sin \theta = a + b \cos (2\theta + y)$

    $\displaystyle \int_{0}^{\pi}(5 \cos^2 \theta + 12 \cos \theta \sin \theta) d \theta = \int_{0}^{\pi}(a + b \cos (2\theta + y)) d \theta$

    $\displaystyle \int_{0}^{\pi}(5 \frac{\cos 2\theta +1}{2} + 6 \sin 2\theta) d \theta = \int_{0}^{\pi}(a + b \cos (2\theta + y)) d \theta$

    $\displaystyle \frac{5}{2} \pi = a \pi$

    $\displaystyle a = \frac{5}{2}$

    $\displaystyle 5 \cos^2 \theta -\frac{5}{2} + 12 \cos \theta \sin \theta = b \cos (2\theta + y)$

    To find b substitute $\displaystyle 5 \cos^2 \theta -\frac{5}{2} = \frac{5}{2} \cos 2\theta$ and $\displaystyle 12 \cos \theta \sin \theta = 6 \sin 2\theta$
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  4. #4
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    i see .....

    so....

    $\displaystyle 5cos^2 \theta + 12 cos \theta sin\theta$

    $\displaystyle =\frac{5}{2}(1+cos2\theta) + 6 sin 2\theta$

    $\displaystyle =\frac{5}{2} + \frac{5}{2}cos2\theta + 6 sin2\theta$

    $\displaystyle =\frac{5}{2} + \frac{13}{2} cos (2\theta - 67.4degree)$

    so.....$\displaystyle a=\frac{5}{2} , b=\frac{13}{2} , y=-67.4degree$

    thanks
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