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Math Help - How solve a trigometric function

  1. #1
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    How solve a trigometric function

    Hey everybody.

    I have the following trig function and I have to simplify it so that it will only have sintheta in it. Can you show me how to do that?:

    9.78049(1+0.005288sinsquaredx - 0.000006sinsquared2x) (m/ssquared)

    I tried pulling out a sinx, but that only seemed to complicate the problem because once I did that had to simplify the sin2x and I just couldn't find a way to do it!

    Is there another way to approach this problem? I've exhausted all sources and my book has the most basic concepts possible!

    Any help is appreciated,
    Stealth
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  2. #2
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    Quote Originally Posted by Stealth View Post
    Hey everybody.

    I have the following trig function and I have to simplify it so that it will only have sintheta in it. Can you show me how to do that?:

    9.78049(1+0.005288sinsquaredx - 0.000006sinsquared2x) (m/ssquared)
    theta?

    a[1 + b\sin^2{x} - c\sin^2(2x)]

    a[1 + b\sin^2{x} - 4c\sin^2{x}\cos^2{x}]

    a[1 + b\sin^2{x} - 4c\sin^2{x}(1 - \sin^2{x})]

    a[1 + b\sin^2{x} - 4c\sin^2{x} + 4c\sin^4{x}]

    a[1 + (b-4c)\sin^2{x} + 4c\sin^4{x}]
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  3. #3
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    Quote Originally Posted by skeeter View Post
    theta?

    a[1 + b\sin^2{x} - c\sin^2(2x)]

    a[1 + b\sin^2{x} - 4c\sin^2{x}\cos^2{x}]

    a[1 + b\sin^2{x} - 4c\sin^2{x}(1 - \sin^2{x})]

    a[1 + b\sin^2{x} - 4c\sin^2{x} + 4c\sin^4{x}]

    a[1 + (b-4c)\sin^2{x} + 4c\sin^4{x}]
    Sorry yes it's x not theta. I replaced it with x to make it easier on my eyes. Um....so I can't simplify it anymore right? I can't drop it down to sinx only?
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  4. #4
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    Quote Originally Posted by Stealth View Post
    Sorry yes it's x not theta. I replaced it with x to make it easier on my eyes. Um....so I can't simplify it anymore right? I can't drop it down to sinx only?
    no
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  5. #5
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    Ok so I'm not the only one that can't solve it anymore...
    Thanks!

    Stealth
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