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    I have no idea how I have messed up this identity proof question

    Prove: Cos4x is identical to 8cos^(4)x-8cos^(2)x+1

    I started with the left hand side, rewriting it as

    Cos(2x+2x)

    I then used the compound formula to obtain

    cos2xcos2x-sin2xsin2x

    Using double angle formulae

    (cos^(2)x-sin^(2)x)^(2) - (2sinxcosx)^2

    Expanding the squared brackets to obtain

    cos^(4)x-2cos^(2)xsin^(2)x+sin^(4)x+4sin^ (2)xcos^(2)x

    Collecting like terms

    sin^(4)x+cos^(4)x+2cos^(2)xsin^( 2)x

    Using the identity sin^(2)x = 1-cos^(2)x and expanding the bracket

    sin^(4)x+cos^(4)x+2cos^(2)x-2cos^(4)x

    Collecting like terms again

    sin^(4)x-cos^(4)x+2cos^(2)x

    Using the identity sin^(4)x = (1-cos^(2)x)^(2)

    1-2cos^(2)x+cos^(4)x-cos^(4)x-2cos^(2)x

    Collecting like terms for the last time cancels out the trig terms, leaving just...

    1

    I know this is a lot of workings, but can somebody please tell me where I've gone wrong?
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    (Original post by jamb97)
    I have no idea how I have messed up this identity proof question

    Prove: Cos4x is identical to 8cos^(4)x-8cos^(2)x+1

    I started with the left hand side, rewriting it as

    Cos(2x+2x)

    I then used the compound formula to obtain

    cos2xcos2x-sin2xsin2x

    Using double angle formulae

    (cos^(2)x-sin^(2)x)^(2) - (2sinxcosx)^2

    Expanding the squared brackets to obtain

    cos^(4)x-2cos^(2)xsin^(2)x+sin^(4)x+4sin^(2)xcos^(2)x
    First error - in red - should be minus.

    Not checked the rest.
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    (Original post by ghostwalker)
    First error - in red - should be minus.

    Not checked the rest.
    But why? Surely if you square that you will get a positive...?

    EDIT: Nevermind, I see what has gone wrong. I misinterpreted my own working... Thank you very much! :P
 
 
 
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