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Math Help - Factoring Trigonometric Expression

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    Senior Member vaironxxrd's Avatar
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    Factoring Trigonometric Expression

    Hello Everyone,
    I have to factor the following expression and I can't figure out the proper procedures to do it .

    I'm assuming that I must take the difference of squares?


    \frac{1}{3} - \frac{2}{3}sin^2(x) + \frac{1}{3}sin^4(x)
    Last edited by vaironxxrd; February 18th 2013 at 06:26 AM.
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    Re: Factoring Trigonometric Expression

    Remove the 1/3 as a common factor and then make use of (a-1)^{2}\equiv a^{2}  - 2a + 1 followed by the trig identity \sin ^{2}A + \cos^{2}A=1.
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    Re: Factoring Trigonometric Expression

    Quote Originally Posted by vaironxxrd View Post
    I have to factor the following expression and I can't figure out the proper procedures to do it .
    \frac{1}{3} - \frac{2}{3}sin^2(x) + \frac{1}{3}sin^4(x)

    \frac{1}{3} - \frac{2}{3}sin^2(x) + \frac{1}{3}sin^4(x)=\frac{(1-sin^2(x))^2}{3}.
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    Senior Member vaironxxrd's Avatar
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    Re: Factoring Trigonometric Expression

    Quote Originally Posted by Plato View Post
    \frac{1}{3} - \frac{2}{3}sin^2(x) + \frac{1}{3}sin^4(x)=\frac{(1-sin^2(x))^2}{3}.
    Plato is the following acceptable?

    \frac{1}{3} - \frac{2}{3}sin^2(x) + \frac{1}{3}sin^4(x) = \frac{1}{3}(1-2sin^2(x)+sin^4(x))
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    Re: Factoring Trigonometric Expression

    Quote Originally Posted by vaironxxrd View Post
    Plato is the following acceptable?

    \frac{1}{3} - \frac{2}{3}sin^2(x) + \frac{1}{3}sin^4(x) = \frac{1}{3}(1-2sin^2(x)+sin^4(x))

    Acceptable to whom?

    How would I know that answer to that?
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