FurtherMaths2020
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https://www.youtube.com/watch?v=VRCNnIJ2cWU

im watching that to help me with this question but it's not really helping. thanks
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RDKGames
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(Original post by FurtherMaths2020)
https://www.youtube.com/watch?v=VRCNnIJ2cWU

im watching that to help me with this question but it's not really helping. thanks
Note that:

\begin{aligned}  C + iS & = (1+\cos \theta + \cos 2 \theta + \ldots + \cos (n-1) \theta ) + i (\sin \theta + \sin 2 \theta + \ldots + \sin (n-1) \theta ) \\ & = 1+(\cos \theta + i\sin \theta) + (\cos 2 \theta + i \sin 2 \theta) + \ldots + (\cos (n-1)\theta + i\sin (n-1)\theta ) \\ & = e^0 + e^{i\theta} + e^{i2\theta} + \ldots + e^{i(n-1)\theta} \\ & = \displaystyle \sum_{k=0}^{n-1} e^{ik\theta} \\ & = \sum_{k=0}^{n-1} (e^{i\theta})^k  \end{aligned}

so far I've shown that the expression is just a geometric sum. You can finish it off from there.
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RDKGames
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(Original post by FurtherMaths2020)
https://www.youtube.com/watch?v=VRCNnIJ2cWU

im watching that to help me with this question but it's not really helping. thanks
Also, that video has nothing to do with this question -- no wonder it's not helping!
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FurtherMaths2020
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(Original post by RDKGames)
Note that:

\begin{aligned}  C + iS & = (1+\cos \theta + \cos 2 \theta + \ldots + \cos (n-1) \theta ) + i (\sin \theta + \sin 2 \theta + \ldots + \sin (n-1) \theta ) \\ & = 1+(\cos \theta + i\sin \theta) + (\cos 2 \theta + i \sin 2 \theta) + \ldots + (\cos (n-1)\theta + i\sin (n-1)\theta ) \\ & = e^0 + e^{i\theta} + e^{i2\theta} + \ldots + e^{i(n-1)\theta} \\ & = \displaystyle \sum_{k=0}^{n-1} e^{ik\theta} \\ & = \sum_{k=0}^{n-1} (e^{i\theta})^k  \end{aligned}

so far I've shown that the expression is just a geometric sum. You can finish it off from there.
so my understanding has taken me to this point. now, I'll try and multiply it through by the conjugate: e^-i(theta)

like it was done in the example ive attached.
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mqb2766
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(Original post by FurtherMaths2020)
so my understanding has taken me to this point. now, I'll try and multiply it through by the conjugate: e^-i(theta)

like it was done in the example ive attached.
Not worked it through, but the obvious thing is to multply top and bottom by the conjugate of the denominator, then equate real and imaginary parts. Should be about right.
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