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Math Help - I don't get this simplification

  1. #1
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    I don't get this simplification

    I can't see how they did this operation.

    thx for the help.
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  2. #2
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    Looks like simple trig identities to me i.e. 2\sin x \cos x = \sin 2x
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  3. #3
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    Quote Originally Posted by konvos View Post
    I can't see how they did this operation.

    thx for the help.
    sin(2x)=2sinxcosx

    -sin^2x+cos^2x=-sin^2x-cos^2x+2cos^2x

    and

    cos^2x+sin^2x=1\Rightarrow\ -\left(sin^2x+cos^2x\right)=-1
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  4. #4
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    Ok..1.sin2(x) + cos2(x) = 1=>sin2(x)=1+cos2(x)....but you have - in front of sin so you can see why -1+cos2(x)
    2.sin(2x) = 2 sin x cos x
    from 1 and 2 =>-sin2(x) + cos2(x)+2 sin x cos x=-1+cos2(x)+sin(2x)
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  5. #5
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    thx for the help everybody

    Quote Originally Posted by Archie Meade View Post
    sin(2x)=2sinxcosx

    -sin^2x+cos^2x=-sin^2x-cos^2x+2cos^2x

    and

    cos^2x+sin^2x=1\Rightarrow\ -\left(sin^2x+cos^2x\right)=-1
    I'd no idea of this identity
    I suppose I should review trig...
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  6. #6
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     -sin^2x + cos^2x = -sin^2x + (cos^2x - cos^2x) + cos^2x =
    = -sin^2x -cos^2x + cos^2x + cos^2x
     = -(sin^2x + cos^2x) + 2cos^2x
     = -1 + 2cos^2x
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  7. #7
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    Quote Originally Posted by konvos View Post
    thx for the help everybody



    I'd no idea of this identity
    I suppose I should review trig...
    No, that's not an identity.
    Just algebraic manipulation.
    You could of course use identities to weave your way to the final line.
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  8. #8
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    Quote Originally Posted by Defunkt View Post
     -sin^2x + cos^2x = -sin^2x + (cos^2x - cos^2x) + cos^2x =
    = -sin^2x -cos^2x + cos^2x + cos^2x
     = -(sin^2x + cos^2x) + 2cos^2x
     = -1 + 2cos^2x
    It's also probably worth noting that \cos^2(x) - \sin^2(x) = \cos(2x) is a standard double angle formula (because I just know that the next question asked by the OP will be how to find y ....)
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