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Math Help - Need help simplifying

  1. #1
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    Need help simplifying

    I'm working with this:

    -\frac{3}{4}t cos(2t) + \frac{3}{16}sin(4t)cos(2t) + cos(2t) + \frac{3}{8}sin^3(2t) + sin(2t)


    ...and I could use some help understanding how this simplifies to some form of:


    Acos(2t) + Bsin(2t) + Ctcos(2t)


    I understand how the sin(4t) breaks down into 2sin(2t)cos(2t), but beyond that, I'm a bit lost.
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  2. #2
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    Re: Need help simplifying

    Quote Originally Posted by Lancet View Post
    I'm working with this:

    -\frac{3}{4}t cos(2t) + \frac{3}{16}sin(4t)cos(2t) + cos(2t) + \frac{3}{8}sin^3(2t) + sin(2t)


    ...and I could use some help understanding how this simplifies to some form of:


    Acos(2t) + Bsin(2t) + Ctcos(2t)


    I understand how the sin(4t) breaks down into 2sin(2t)cos(2t), but beyond that, I'm a bit lost.
    Dear Lancet,

    -\frac{3t}{4}\cos(2t) + \frac{3}{16}\sin(4t)\cos(2t) + \cos(2t) + \frac{3}{8}\sin^{3}(2t) +\sin(2t)


    =-\frac{3t}{4}\cos(2t) + \frac{3}{8}\sin (2t)\cos^{2}(2t) + \cos(2t) + \frac{3}{8}\sin^{3}(2t) + \sin(2t)

    =-\frac{3t}{4}\cos(2t) + \frac{3}{8}\sin (2t)(1-\sin^{2}(2t)) + \cos(2t) + \frac{3}{8}\sin^{3}(2t) + \sin(2t)

    =-\frac{3t}{4}\cos(2t) + \frac{3}{8}\sin (2t)-\frac{3}{8}\sin^{3}(2t) + \cos(2t) + \frac{3}{8}\sin^{3}(2t) + \sin(2t)

    =-\frac{3t}{4}\cos(2t) + \frac{3}{8}\sin (2t)+ \cos(2t) +\sin(2t)

    =-\frac{3t}{4}\cos(2t)+\frac{11}{8}\sin (2t)+\cos(2t)

    Therefore, A=1~,~B=\frac{11}{8}\mbox{ and }C=-\frac{3}{4}
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  3. #3
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    Re: Need help simplifying

    Ah! Now I understand. Thank you for taking the time to explain it to me.
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