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Thread: trigonometric simplification

  1. #1
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    trigonometric simplification

    Hi,

    I seem to have forgotten much about trig. identities and so on.

    How did the authors reduce the following:

    $\displaystyle 32\cos^2{t}\sin^2{t}$

    $\displaystyle = 8\sin^2{2t}$

    $\displaystyle = 4-4\cos{4t}$
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  2. #2
    Like a stone-audioslave ADARSH's Avatar
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    The following two identities were used

    2sin(x)cos(x) =sin(2x)

    sin^2(y) = (1-cos(2y))/2
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  3. #3
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    Trig Identities

    Hello scorpion007
    Quote Originally Posted by scorpion007 View Post
    Hi,

    I seem to have forgotten much about trig. identities and so on.

    How did the authors reduce the following:

    $\displaystyle 32\cos^2{t}\sin^2{t}$

    $\displaystyle = 8\sin^2{2t}$

    $\displaystyle = 4-4\cos{4t}$
    First, note the identity: $\displaystyle \sin2t = 2\sin t\cos t$

    $\displaystyle \Rightarrow 8\sin^22t = 8(2\sin t \cos t)^2 = 32\sin^2t\cos^2t$

    and then the identity $\displaystyle \cos 2t = 1 - 2\sin^2t$

    $\displaystyle \Rightarrow \sin^2t = \tfrac12(1-\cos 2t)$

    $\displaystyle \Rightarrow \sin^22t= \tfrac12(1-\cos 4t)$

    $\displaystyle \Rightarrow 8\sin^22t = 4(1 -\cos 4t)$

    Grandad
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  4. #4
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    Ah, thank you both very much for the explanation!
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