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Thread: Trigometric identity problem.

  1. #1
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    Trigometric identity problem.

    prove
    $\displaystyle sin^4t=3/8-1/2*cos2t+1/8*cos4t$
    get rid of the fraction so $\displaystyle 3-4cos2t+8cos4t$
    $\displaystyle 3-4(1-2sin^2t)+8(1-2sin^2t)^2$
    $\displaystyle 7+8sin^2t-32sin^2t+32sin^4t$
    $\displaystyle 7-24sin^2t+32sin^4t$
    this looks nothing like $\displaystyle sin^4$...
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  2. #2
    Super Member 11rdc11's Avatar
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    Quote Originally Posted by dorkymichelle View Post
    prove
    $\displaystyle sin^4t=3/8-1/2*cos2t+1/8*cos4t$
    get rid of the fraction so $\displaystyle 3-4cos2t+8cos4t$
    $\displaystyle 3-4(1-2sin^2t)+8(1-2sin^2t)^2$
    $\displaystyle 7+8sin^2t-32sin^2t+32sin^4t$
    $\displaystyle 7-24sin^2t+32sin^4t$
    this looks nothing like $\displaystyle sin^4$...
    Remember that

    $\displaystyle \sin^4{t} = (\sin^2{t})^2 = \bigg(\frac{1-\cos{2t}}{2}\bigg)^2$


    which expands to

    $\displaystyle \frac{1}{4}\bigg(1-2\cos{2t}+\cos^2{2t}\bigg)$

    Now expand

    $\displaystyle \cos^2{2t} = \frac{\cos{4t} +1}{2}$

    plug that back in and you get

    $\displaystyle \frac{1}{4}\bigg(1-2\cos{2t}+\frac{\cos{4t} +1}{2}\bigg)$

    simplify that and you get your identity
    Last edited by 11rdc11; Dec 9th 2009 at 11:55 PM.
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  3. #3
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    isnt $\displaystyle sin^4 = (sin^2)(sin^2)$?
    so wouldn't it be $\displaystyle (1-cos^2)^2$?
    Where did that formula come from?
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  4. #4
    Super Member 11rdc11's Avatar
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    Quote Originally Posted by dorkymichelle View Post
    isnt $\displaystyle sin^4 = (sin^2)(sin^2)$?
    so wouldn't it be $\displaystyle (1-cos^2)^2$?
    Where did that formula come from?

    $\displaystyle \cos{2t} = \cos^2{t} - \sin^2{t}$

    $\displaystyle \cos{2t} = (1-sin^2{t}) - \sin^2{t}$

    $\displaystyle \cos{2t} = 1-2sin^2{t}$

    $\displaystyle \cos{2t}-1 = -2sin^2{t}$

    $\displaystyle \frac{\cos{2t}-1}{-2} = sin^2{t}$

    $\displaystyle \frac{1-\cos{2t}}{2} = sin^2{t}$
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