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    Hi again.

    Right so the question is

    By using the identity cos 2A = 1 - 2 sin^2 A show that cos 4x = 1 - 8 sin^2 x cos^2 x

    I have put

    cos 4x ---> cos (2x + 2x)

    = 1 - 2 sin^2 x + 1 - 2 sin^2 x

    = 2 - 4 sin^2 x

    So where the hell have they got their 8 and cos^2 x from?? I tried using the identities for cos 2x that have cos^2x in but it hasn't got me much closer to the answer they want.

    TSR mathematicians, help me please!!! And thank you in advance, you guys have helped me a lot this weekend
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    cos4x = 1 - 2sin^2(2x) is the first line they were looking for.

    Can you see what to do next? Work on the sin2x term.
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    (Original post by HaNzY)
    I have put

    cos 4x ---> cos (2x + 2x)

    = 1 - 2 sin^2 x + 1 - 2 sin^2 x
    no, you have to substitute the "x"s with "2x"s

    1 - 2sin^2(2x)
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    (Original post by HaNzY)
    cos 4x ---> cos (2x + 2x)

    = 1 - 2 sin^2 x + 1 - 2 sin^2 x
    Eurgh, I just realised what you did here. You've assumed that cos(2x + 2x) = cos(2x) + cos(2x) which is very, very bad and big school boy error.

    cos(2x + 2x) = cos(2x)cos(2x) - sin(2x)sin(2x)
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    Tip:

     sin^2(2x) = 1-cos^2(2x) and you already have an expression for  cos(2x) in the question...
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    (Original post by Swayum)
    cos4x = 1 - 2sin^2(2x) is the first line they were looking for.

    Can you see what to do next? Work on the sin2x term.
    I did this (which came out with the right answer!!)

    1 - 2 (2 sinx cosx)^2

    1 - 2 x 4 sin^2x cos^2x

    =1 - 8 sin^2x cos^2x !!

    Is this right?? Or have I forged a connection lol?
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    Yup, that's right
 
 
 
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