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Math Help - Prooving and Identity

  1. #1
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    Prooving and Identity

    Hi, I'm struggling to understand how to prove this:

    (sin^4) T = ( 3 - 4 cos 2 T + cos 4 T ) / 8

    (where T is theta)

    I start on the left side and "break" this up into a square:

    ( sin^2 T) ^ 2

    From here, my book does the following:

    ( 1 - cos 2 T / 2 ) ^ 2

    Why?

    I know sin^2 is also = 1 - cos ^ 2
    How do I know to use the double angle identity?? It doesn't seem to apply if I'm just working with sin^2 T?

    Thanks!
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  2. #2
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    One of the ways to write the Double Angle Identiy for Cosine is:

    \cos(2\theta)=1-2sin^{2}\theta\Rightarrow
    2sin^{2}\theta=1-\cos(2\theta)\Rightarrow
    sin^{2}\theta=\frac{1-\cos(2\theta)}{2}

    You can rearrange this to come to the step your book has taken.
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  3. #3
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    Hello, Snowcrash!

    Prove: . \sin^4\!\theta\:=\: \frac{<br />
3 - 4\cos2\theta + \cos4\theta}{8}

    I start on the left side and "break" this up into a square: .  (\sin^2\!\theta)^ 2


    From here, my book does the following: . \left(\frac{1 - \cos2\theta}{2}\right)^ 2
    Why?

    I know: . \sin^2\!\theta \:=\:1 - \cos^2 \!\theta

    How do I know to use the double angle identity?

    How else are you going to change your (1-\cos^2\!\theta)^2
    . . into an expression with . \cos2\theta\,\text{ and }\,\cos4\theta ?


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