Answer: (a) \(1+z+z^2+\ldots+z^{n-1}=\dfrac{z^n-1}{z-1}\), for \(z\neq 1\).

(b) \(1+\cos\theta+\cos2\theta+\ldots+\cos(n-1)\theta=\dfrac12\left(1-\cos n\theta+\dfrac{\sin n\theta\sin\theta}{1-\cos\theta}\right)\).

(c) \(\displaystyle \int_0^1 \cos x\,dx\lt \dfrac{1}{2n}\left(1-\cos1+\dfrac{\sin1\sin\frac1n}{1-\cos\frac1n}\right)\).

(d) A lower bound is

\(\displaystyle \int_0^1 \cos x\,dx\gt \dfrac{1}{2n}\left(\cos1-1+\dfrac{\sin1\sin\frac1n}{1-\cos\frac1n}\right)\).

(a) This is a finite geometric series with first term \(1\), common ratio \(z\), and \(n\) terms, so

\(1+z+z^2+\ldots+z^{n-1}=\dfrac{1-z^n}{1-z}=\dfrac{z^n-1}{z-1}\).

(b) Let \(z=\cos\theta+\mathrm{i}\sin\theta\). Then by de Moivre's theorem,

\(z^n=\cos n\theta+\mathrm{i}\sin n\theta\).

From part (a),

\(1+z+z^2+\ldots+z^{n-1}=\dfrac{z^n-1}{z-1}=\dfrac{\cos n\theta-1+\mathrm{i}\sin n\theta}{\cos\theta-1+\mathrm{i}\sin\theta}\).

Multiply numerator and denominator by the conjugate \(\cos\theta-1-\mathrm{i}\sin\theta\):

\(\dfrac{z^n-1}{z-1}=\dfrac{(\cos n\theta-1+\mathrm{i}\sin n\theta)(\cos\theta-1-\mathrm{i}\sin\theta)}{(\cos\theta-1)^2+\sin^2\theta}\).

The denominator is

\((\cos\theta-1)^2+\sin^2\theta=\cos^2\theta-2\cos\theta+1+\sin^2\theta=2(1-\cos\theta)\).

The real part of the numerator is

\(\cos n\theta\cos\theta+\sin n\theta\sin\theta-\cos n\theta-\cos\theta+1\).

Hence

\(\operatorname{Re}\left(\dfrac{z^n-1}{z-1}\right)=\dfrac{\cos n\theta\cos\theta+\sin n\theta\sin\theta-\cos n\theta-\cos\theta+1}{2(1-\cos\theta)}\).

Now separate the final two terms:

\(=\dfrac{\cos n\theta\cos\theta+\sin n\theta\sin\theta-\cos n\theta}{2(1-\cos\theta)}+\dfrac{1-\cos\theta}{2(1-\cos\theta)}\)

\(=\dfrac{\cos n\theta(\cos\theta-1)+\sin n\theta\sin\theta}{2(1-\cos\theta)}+\dfrac12\).

Since \(\cos\theta-1=-(1-\cos\theta)\), this becomes

\(\operatorname{Re}\left(\dfrac{z^n-1}{z-1}\right)=\dfrac12\left(1-\cos n\theta+\dfrac{\sin n\theta\sin\theta}{1-\cos\theta}\right)\).

The real part of the left-hand side is

\(1+\cos\theta+\cos2\theta+\ldots+\cos(n-1)\theta\),

so

\(1+\cos\theta+\cos2\theta+\ldots+\cos(n-1)\theta=\dfrac12\left(1-\cos n\theta+\dfrac{\sin n\theta\sin\theta}{1-\cos\theta}\right)\).

(c) On \([0,1]\), \(\cos x\) is decreasing, so rectangles of width \(\dfrac1n\) taken from the left endpoints lie above the curve. Therefore

\(\int_0^1 \cos x\,dx\lt \dfrac1n+\dfrac1n\cos\dfrac1n+\dfrac1n\cos\dfrac2n+\ldots+\dfrac1n\cos\dfrac{n-1}{n}\)

\(=\dfrac1n\left(1+\cos\dfrac1n+\cos\dfrac2n+\ldots+\cos\dfrac{n-1}{n}\right).\)

Using the result from part (b) with \(\theta=\dfrac1n\),

\(1+\cos\dfrac1n+\cos\dfrac2n+\ldots+\cos\dfrac{n-1}{n}=\dfrac12\left(1-\cos\dfrac{n}{n}+\dfrac{\sin\dfrac{n}{n}\sin\dfrac1n}{1-\cos\dfrac1n}\right).\)

Since \(\dfrac{n}{n}=1\),

\(\int_0^1 \cos x\,dx\lt \dfrac{1}{2n}\left(1-\cos1+\dfrac{\sin1\sin\dfrac1n}{1-\cos\dfrac1n}\right).\)

(d) For a lower bound, use right-endpoint rectangles. Since \(\cos x\) is decreasing on \([0,1]\), these lie below the curve, so

\(\int_0^1 \cos x\,dx\gt \dfrac1n\cos\dfrac1n+\dfrac1n\cos\dfrac2n+\ldots+\dfrac1n\cos\dfrac{n}{n}.\)

Now

\(\cos\dfrac1n+\cos\dfrac2n+\ldots+\cos\dfrac{n}{n}\)

\(=\left(1+\cos\dfrac1n+\cos\dfrac2n+\ldots+\cos\dfrac{n-1}{n}\right)+\cos1-1.\)

Substitute the expression from part (c):

\(=\dfrac12\left(1-\cos1+\dfrac{\sin1\sin\dfrac1n}{1-\cos\dfrac1n}\right)+\cos1-1\)

\(=\dfrac12\left(\cos1-1+\dfrac{\sin1\sin\dfrac1n}{1-\cos\dfrac1n}\right).\)

Therefore

\(\int_0^1 \cos x\,dx\gt \dfrac{1}{2n}\left(\cos1-1+\dfrac{\sin1\sin\dfrac1n}{1-\cos\dfrac1n}\right).\)