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Math Help - Trig factoring

  1. #1
    Junior Member Greymalkin's Avatar
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    Trig factoring

    Through what operation(s) does

    \frac{1}{5}[sin(4x)+1]^\frac{5}{2}-\frac{1}{14}[sin(4x)+1]^\frac{7}{2}+C

    become

    \frac{1}{70}[9-5sin(4x))(sin(4x)+1]^\frac{5}{2}+C
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  2. #2
    MHF Contributor MarkFL's Avatar
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    Re: Trig factoring

    Observe that the initial expression may be written:

    \frac{14}{70}\left(\sin(4x)+1 \right)^{\frac{5}{2}}-\frac{5}{70}\left(\sin(4x)+1 \right)\left(\sin(4x)+1 \right)^{\frac{5}{2}}+C

    Now factor:

    \frac{1}{70}\left(\sin(4x)+1 \right)^{\frac{5}{2}}\left(14-5(\sin(4x)+1) \right)+C=

    \frac{1}{70}\left(9-5\sin(4x) \right)\left(\sin(4x)+1 \right)^{\frac{5}{2}}+C
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  3. #3
    Junior Member Greymalkin's Avatar
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    Re: Trig factoring

    Quote Originally Posted by MarkFL2 View Post
    Observe that the initial expression may be written:

    \frac{14}{70}\left(\sin(4x)+1 \right)^{\frac{5}{2}}-\frac{5}{70}\left(\sin(4x)+1 \right)\left(\sin(4x)+1 \right)^{\frac{5}{2}}+C
    I understand that you remove a factor of sin4x+1 from the exponent, but where do the \frac{14}{70} and the \frac{5}{70} come from?

    Now factor:

    Quote Originally Posted by MarkFL2 View Post
    \frac{1}{70}\left(\sin(4x)+1 \right)^{\frac{5}{2}}\left(14-5(\sin(4x)+1) \right)+C=

    \frac{1}{70}\left(9-5\sin(4x) \right)\left(\sin(4x)+1 \right)^{\frac{5}{2}}+C
    What are you doing with the 14 and the 9?
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  4. #4
    MHF Contributor MarkFL's Avatar
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    Re: Trig factoring

    I am rewriting the given fractions using a common denominator of 70.

    Once you factor the 1/70 out, then you are left with the numerators of 14 and 5. From there it is a matter of distributing and collecting like terms.
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