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    If C = 1+\cos\theta+...+\cos(n-1)\theta and S = \sin\theta+...+\sin(n-1)\theta.

    prove that C=\frac{\sin\frac{n\theta}{2}}{\  sin\frac{\theta}{2}} \cos\frac{(n-1)\theta}{2} and S =\frac{\sin\frac{n\theta}{2}}{\s  in\frac{\theta}{2}}\sin\frac{(n-1)\theta}{2} if  \theta\neq2k\pi, k\in\mathbb{Z}

    Attempt

    C+iS = 1+(\cos\theta+i\sin\theta)+...+(  \cos(n-1)\theta+i\sin(n-1)\theta)

    =1+e^{i\theta}+...+e^{i(n-1)\theta}

    =1+z+...+z^{n-1} where z=e^{i\theta}

    =\frac{1-z^n}{1-z} if z\neq1

    =\frac{1-e^{in\theta}}{1-e^{i\theta}}=\frac{e^\frac{in \theta}{2}(e^\frac{-in\theta}{2})-e^\frac{in\theta}{2}}{e\frac{i \theta}{2}(e\frac{-i\theta}{2})-e\frac{i\theta}{2}}

    Is this correct so far for the derivation? How can I prove that the real parts are equal to the complex parts?

    Thanks.
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    I would guess that you first rationalise the fraction so the denominator is real by multiplying by the complex conjugate, then just equate the real and imaginary parts
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    (Original post by Darth_Narwhale)
    I would guess that you first rationalise the fraction so the denominator is real by multiplying by the complex conjugate, then just equate the real and imaginary parts
    but is it derived correctly?
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    (Original post by AishaGirl)
    but is it derived correctly?
    looks fine
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    (Original post by AishaGirl)
    =\frac{1-e^{in\theta}}{1-e^{i\theta}}=\frac{e^\frac{in \theta}{2}(e^\frac{-in\theta}{2})-e^\frac{in\theta}{2}}{e\frac{i \theta}{2}(e\frac{-i\theta}{2})-e\frac{i\theta}{2}}
    Did you mean

    \displaystyle =\frac{1-e^{in\theta}}{1-e^{i\theta}}=\frac{e^\frac{in \theta}{2}(e^\frac{-in\theta}{2}-e^\frac{in\theta}{2})}{e\frac{i \theta}{2}(e\frac{-i\theta}{2}-e\frac{i\theta}{2})} (note brackets)?

    (Original post by Darth_Narwhale)
    I would guess that you first rationalise the fraction so the denominator is real by multiplying by the complex conjugate, then just equate the real and imaginary parts
    At this point, it works out much easier if you go:

    \displaystyle = \frac{e^\frac{i(n-1) \theta}{2}(e^\frac{-in\theta}{2}-e^\frac{in\theta}{2})}{e\frac{-i\theta}{2}-e\frac{i\theta}{2}}

    Now, rather mulitiplying by the complex conjugate, notice that the bottom is now -2i \sin(\theta /2) and so you can make it real simply by multipllying by i instead.
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    (Original post by DFranklin)
    Did you mean

    \displaystyle =\frac{1-e^{in\theta}}{1-e^{i\theta}}=\frac{e^\frac{in \theta}{2}(e^\frac{-in\theta}{2}-e^\frac{in\theta}{2})}{e\frac{i \theta}{2}(e\frac{-i\theta}{2}-e\frac{i\theta}{2})} (note brackets)?

    At this point, it works out much easier if you go:

    \displaystyle = \frac{e^\frac{i(n-1) \theta}{2}(e^\frac{-in\theta}{2}-e^\frac{in\theta}{2})}{e\frac{-i\theta}{2}-e\frac{i\theta}{2}}

    Now, rather mulitiplying by the complex conjugate, notice that the bottom is now -2i \sin(\theta /2) and so you can make it real simply by multipllying by i instead.
    Ah yeah small mistake with the brackets. Thanks DFranklin.
 
 
 
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