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    Hi!

    Im finding trigonometric identities relatively hard, and Im wondering if anyone can help me solve the question below:

    (X replaces THETA)


    Code:
    Show that 8Cos^4X = cos4X + 4cos2X + 3
    Thanks!
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    Hi!

    Im finding trigonometric identities relatively hard, and Im wondering if anyone can help me solve the question below:

    (X replaces THETA)
    Use this identity multiple times

     cos^2(\theta) = \frac{cos(2\theta)+1}{2}
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    (Original post by spread_logic_not_hate)
    Use this identity multiple times

     cos^2(\theta) = \frac{cos(2\theta)+1}{2}
    I can't fault logic's method, but you did specify FP2.

    Part of the FP2 idea is to be able to use De Moivre to expand cos(nx) or sin(nx). You then get some polynomial in cos and/or sin to play with.
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    (Original post by ian.slater)
    I can't fault logic's method, but you did specify FP2.

    Part of the FP2 idea is to be able to use De Moivre to expand cos(nx) or sin(nx). You then get some polynomial in cos and/or sin to play with.
    Yeh!, I need to use De Moivre.

    Can you help?
    Thanks!
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    OK

    exp(ix) = cos(x) + isin(x) by definition

    so exp(nix) = (cos(x) + isin(x))^n

    but also exp(nix) = cos(nx) + isin(nx)

    If you expand say cos(4x) + isin(4x) = (cos(x) + isin(x))^4

    you equate real parts to get cos(4x) and imag parts to get sin(4x).

    Then the examiner dreams up ways to get you to use that
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    Yeh!, I need to use De Moivre.

    Can you help?
    Thanks!
    Ah sorry did not know that. Ok then, De Moivre's theorem is

     (cos x + i sin x)^n = cos(nx) + i sin (nx)

    In your case i'd write this

     (cos x + i sin x)^4 =  [(cos x + i sin x)^2]^2 = cos(4x) + i sin (4x)

    Now expand the LHS and collect all the real terms, which will be equal to cos(4x)...
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    In the question, I'd use DM to expand cos(4x) (remembering that sin^2 = 1-cos^2) and also cos(2x) (although you could just quote that). Work out the RHS and hope when the smoke clears you're left with the LHS
 
 
 
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