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Math Help - Application - Rewriting A Trigonometric Function

  1. #1
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    Application - Rewriting A Trigonometric Function

    Acceleration due to gravity, "g", is considered to be 9.8 \frac {m}{s^2}. However, changes in latitude alter "g" according to the following model:

    g \approx 9.78049(1+.005288\sin^2{\theta} - .000006\sin^2{2\theta})\frac {m}{s^2}.

    a) Rewrite "g" in terms of power \sin{\theta} only.
    I know some of the trigonometric identities but I keep getting stuck on rewriting it in terms or \sin{\theta}. Help?

    b) Find the latitude of Denver, Colorado and determite "g".
    I'll be able to do this once I finish part (a).

    (c) Find the percent change between the value calculated in part (b) above and the standard value of 9.8.

    I'll be able to do this once I finish part (a) and (b).
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  2. #2
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    Quote Originally Posted by chrozer View Post
    Acceleration due to gravity, "g", is considered to be 9.8 \frac {m}{s^2}. However, changes in latitude alter "g" according to the following model:

    g \approx 9.78049(1+.005288\sin^2{\theta} - .000006\sin^2{2\theta})\frac {m}{s^2}.

    a) Rewrite "g" in terms of power \sin{\theta} only.
    I know some of the trigonometric identities but I keep getting stuck on rewriting it in terms or \sin{\theta}. Help?

    b) Find the latitude of Denver, Colorado and determite "g".
    I'll be able to do this once I finish part (a).

    (c) Find the percent change between the value calculated in part (b) above and the standard value of 9.8.

    I'll be able to do this once I finish part (a) and (b).
    Change  sin(2\theta) = 2sin(\theta)cos(2\theta)

    and then use  cos^{2}(\theta) = 1 - sin^{2}(\theta)
    Then it will be in terms of  sin(\theta)
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  3. #3
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    Quote Originally Posted by bobby View Post
    Change  sin(2\theta) = 2sin(\theta)cos(2\theta)

    and then use  cos^{2}(\theta) = 1 - sin^{2}(\theta)
    Then it will be in terms of  sin(\theta)
    But isn't \sin^2{\theta} not in terms of  sin(\theta) ?
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  4. #4
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    Quote Originally Posted by chrozer View Post
    But isn't \sin^2{\theta} not in terms of  sin(\theta) ?
     \sin^2{\theta} is in terms of \sin(\theta)

    It's just the  \theta bit that matters.
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  5. #5
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    Quote Originally Posted by bobby View Post
     \sin^2{\theta} is in terms of \sin(\theta)

    It's just the  \theta bit that matters.
    Ok I see. Thanks alot.
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