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Thread: 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 $\displaystyle 9.8 \frac {m}{s^2}$. However, changes in latitude alter "g" according to the following model:

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

    a) Rewrite "g" in terms of power $\displaystyle \sin{\theta}$ only.
    I know some of the trigonometric identities but I keep getting stuck on rewriting it in terms or $\displaystyle \sin{\theta}$. 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 $\displaystyle 9.8 \frac {m}{s^2}$. However, changes in latitude alter "g" according to the following model:

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

    a) Rewrite "g" in terms of power $\displaystyle \sin{\theta}$ only.
    I know some of the trigonometric identities but I keep getting stuck on rewriting it in terms or $\displaystyle \sin{\theta}$. 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 $\displaystyle sin(2\theta) = 2sin(\theta)cos(2\theta) $

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

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

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

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