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    Hi, I'm stuck on the first question (1a) of excersize 7C. The question is:
    Show that:
     1+4cos2x + 6cos4x + 4cos6x + cos8x = 16cos4x  cos^4 x
    The examples on the pages before are all about sums of infinite geometric series which I understood, but I'm not really sure how this question relates to the example in any way.

    Help would be appreciated
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    Well you can write the LHS as \text{Re}(1+4e^{2ix} + 6e^{4ix} + 4e^{6ix} + e^{8ix}). As a hint, think Pascal's triangle... how can you factorise this expression?
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    Ah ye, I'm thinking its something to the power 4, because of those binomial coefficients. Is it (1+e^2ix)^4?
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    (Original post by jamie092)
    Ah ye, I'm thinking its something to the power 4, because of those binomial coefficients. Is it (1+e^2ix)^4?
    Yup, so you have to find the real part of (1+e^{2ix})^4.
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    (Original post by nuodai)
    Yup, so you have to find the real part of (1+e^{2ix})^4.
    Surely thats just (1+cos2x)^4?
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    (Original post by jamie092)
    Surely thats just (1+cos2x)^4?
    Nope, in general it's not true that \text{Re}(zw) = \text{Re}(z) \text{Re}(w) so you can't just take the "Re" inside the brackets.
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    (Original post by nuodai)
    Nope, in general it's not true that \text{Re}(zw) = \text{Re}(z) \text{Re}(w) so you can't just take the "Re" inside the brackets.
    Oh ye ofcourse ;<
    Well now I'm stumped =(
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    (Original post by jamie092)
    Oh ye ofcourse ;<
    Well now I'm stumped =(
    Well if z=x+iy then z^4 = x^4 + 4ix^3y - 6x^2y^2 - 4ixy^3 + y^4. You'll notice that the terms with even powers of y are real, and the terms with odd powers of y are imaginary (because of how the powers of i work). So set z=1+e^{2i\theta} = (1+\cos 2\theta) + i\sin 2\theta \equiv x+iy and expand as above. A bit of tedious algebraic manipulation should give you the right answer.
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    (Original post by nuodai)
    Well if z=x+iy then z^4 = x^4 + 4ix^3y - 6x^2y^2 - 4ixy^3 + y^4. You'll notice that the terms with even powers of y are real, and the terms with odd powers of y are imaginary (because of how the powers of i work). So set z=1+e^{2i\theta} = (1+\cos 2\theta) + i\sin 2\theta \equiv x+iy and expand as above. A bit of tedious algebraic manipulation should give you the right answer.
    ok thanks I'll give it a go
 
 
 
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