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    (cosx - sinx)2 in terms of sin2x
    I expanded the brackets and got the answer to just be cos2x, I cant seem to think how I could express it in terms of sin2x.

    cos4x - sin4x in terms of cos2x
    I got the answer to be cos2(2x) but the book says the answer is just cos2x
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    (Original post by znx)
    Express:

    (cosx - sinx)2 in terms of sin2x
    I expanded the brackets and got the answer to just be cos2x, I cant seem to think how I could express it in terms of sin2x.

    cos4x - sin4x in terms of cos2x
    I got the answer to be cos2(2x) but the book says the answer is just cos2x
    Not quite correct. Expansion is \cos^2x - 2\cos x \sin x + \sin^2 x

    Secondly, note that \cos^4 x - \sin^4 x \neq (\cos^2 x - \sin^2 x )^2. Rather, \cos^4 x - \sin^4 x=(\cos^2x - \sin^2 x )(\cos^2 x + \sin^2 x)
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    (Original post by RDKGames)
    Not quite correct. Expansion is \cos^2x - 2\cos x \sin x + \sin^2 x

    Secondly, note that \cos^4 x - \sin^4 x \neq (\cos^2 x - \sin^2 x )^2. Rather, \cos^4 x - \sin^4 x=(\cos^2x - \sin^2 x )(\cos^2 x + \sin^2 x)
    so (cos2x - sin2x)(cos2x + sin2x) = (cos(2x))(cos(2x)) = cos2(2x)?
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    (Original post by znx)
    so (cos2x - sin2x)(cos2x + sin2x) = (cos(2x))(cos(2x)) = cos2(2x)?
    \cos^2x + \sin^2 x \neq \cos(2x)
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    (cosx - sinx)2 in terms of sin2x

    cos4x - sin4x in terms of cos2x

    a) You'd get cos^2(x) + sin^2(x) - 2sin(x)cos(x). What do you notice about cos^2(x) + sin^2(x) and the -2sin(x)cos(x). What identities are these?

    b) cos4x - sin4x = [cos^2(x) + sin^2(x)][cos^2(x)-sin^2(x]. Look at the 2 brackets. You know the first equals 1 and the second equals cos(2x). So 1cos(2x) = ... ?
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    (Original post by RDKGames)
    \cos^2x + \sin^2 x \neq \cos(2x)
    Oh yeah its just 1, so the answer would just be cos2x, thanks again mate
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    (Original post by RDKGames)
    \cos^2x + \sin^2 x \neq \cos(2x)
    How do you do that fancy text again?
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    (Original post by Bulletzone)
    How do you do that fancy text again?
    https://www.thestudentroom.co.uk/help/latex
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    (Original post by thekidwhogames)
    Express:

    (cosx - sinx)2 in terms of sin2x

    cos4x - sin4x in terms of cos2x

    a) You'd get cos^2(x) + sin^2(x) - 2sin(x)cos(x). What do you notice about cos^2(x) + sin^2(x) and the -2sin(x)cos(x). What identities are these?

    b) cos4x - sin4x = [cos^2(x) + sin^2(x)][cos^2(x)-sin^2(x]. Look at the 2 brackets. You know the first equals 1 and the second equals cos(2x). So 1cos(2x) = ... ?
    Thank you mate, managed to get both right answers
 
 
 
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