Answer:
The sum of the geometric series is \(z+z^2+z^3+\cdots+z^n=\dfrac{z(z^n-1)}{z-1}\) for \(z\neq 1\).
Hence, with \(z=\cos\theta+i\sin\theta=e^{i\theta}\), the required sum of cosines is the real part of this expression. Therefore
\(\cos\theta+\cos2\theta+\cdots+\cos n\theta=\Re\left(\dfrac{z(z^n-1)}{z-1}\right)=\dfrac{\sin\frac12 n\theta}{\sin\frac12\theta}\cos\frac12(n+1)\theta\),
provided \(\sin\frac12\theta\neq 0\).
Let \(S=z+z^2+z^3+\cdots+z^n\). This is a geometric series with first term \(z\) and common ratio \(z\), so
\(S=\dfrac{z(z^n-1)}{z-1}\), for \(z\neq 1\).
Now let \(z=\cos\theta+i\sin\theta=e^{i\theta}\). Then
\(\cos\theta+\cos2\theta+\cdots+\cos n\theta=\Re(S)=\Re\left(\dfrac{z(z^n-1)}{z-1}\right).\)
Write \(z=e^{i\theta}\). Then
\(\Re\left(\dfrac{z(z^n-1)}{z-1}\right)=\Re\left(\dfrac{e^{i\theta}(e^{in\theta}-1)}{e^{i\theta}-1}\right).\)
To simplify the denominator, multiply top and bottom by \(e^{-i\theta/2}\):
\(\dfrac{e^{i\theta}(e^{in\theta}-1)}{e^{i\theta}-1}=\dfrac{e^{i(n+1/2)\theta}-e^{i\theta/2}}{e^{i\theta/2}-e^{-i\theta/2}}=\dfrac{e^{i(n+1/2)\theta}-e^{i\theta/2}}{2i\sin(\theta/2)}.\)
Taking the real part gives
\(\cos\theta+\cos2\theta+\cdots+\cos n\theta=\dfrac{\sin\left(\left(n+\tfrac12\right)\theta\right)-\sin\left(\tfrac12\theta\right)}{2\sin\left(\tfrac12\theta\right)}.\)
Now use \(\sin A-\sin B=2\cos\left(\dfrac{A+B}{2}\right)\sin\left(\dfrac{A-B}{2}\right)\) with \(A=\left(n+\tfrac12\right)\theta\) and \(B=\tfrac12\theta\):
\(\sin\left(\left(n+\tfrac12\right)\theta\right)-\sin\left(\tfrac12\theta\right)=2\cos\left(\tfrac12(n+1)\theta\right)\sin\left(\tfrac12 n\theta\right).\)
So
\(\cos\theta+\cos2\theta+\cdots+\cos n\theta=\dfrac{\sin\frac12 n\theta}{\sin\frac12\theta}\cos\frac12(n+1)\theta,\)
as required, where \(\sin\frac12\theta\neq 0\).
