Sums of Series
Package a trigonometric series as the real or imaginary part of a geometric series in $e^{i\theta}$, sum the geometric series, and split the result back apart.
Setting up a series with $z=e^{i\theta}$
Many trigonometric series can be summed by recognising them as the real or imaginary part of a geometric series in $z=e^{i\theta}$ (or, more generally, $z=re^{i\theta}$). The strategy has four steps:
- Write the target series $C$ (cosines) and its imaginary twin $S$ (sines) together as $C+iS$.
- Recognise $C+iS$ as the sum of a geometric series with common ratio $z=e^{i\theta}$ (or similar), and sum it using the standard geometric series formula.
- Manipulate the resulting closed form algebraically (often multiplying top and bottom by a conjugate-like factor) until it separates cleanly into a real part and an imaginary part.
- Read off $C=\operatorname{Re}(C+iS)$ and $S=\operatorname{Im}(C+iS)$.
For a geometric series with first term $a$ and common ratio $r\neq1$, the sum of the first $n$ terms is $\dfrac{a(1-r^n)}{1-r}$, and if $|r|<1$ the infinite sum is $\dfrac{a}{1-r}$. When $r=e^{i\theta}$, $|r|=1$ so the infinite-sum formula never applies directly — only the finite formula is used for finite trigonometric series in this lesson.
Let $C=1+\cos\theta+\cdots+\cos(n-1)\theta$ and $S'=0+\sin\theta+\cdots+\sin(n-1)\theta$ (matching term-by-term, including a leading $\sin0\theta=0$ term for symmetry). Then
$C+iS'=1+(\cos\theta+i\sin\theta)+(\cos2\theta+i\sin2\theta)+\cdots+(\cos(n-1)\theta+i\sin(n-1)\theta)=1+z+z^2+\cdots+z^{n-1}$
where $z=e^{i\theta}$, using De Moivre's theorem to write $z^k=\cos k\theta+i\sin k\theta$. This is a geometric series with first term $1$, ratio $z$, and $n$ terms, so
$$C+iS'=\frac{1(1-z^n)}{1-z}=\frac{1-e^{in\theta}}{1-e^{i\theta}}$$
Taking real and imaginary parts gives $C=\operatorname{Re}\left(\dfrac{1-e^{in\theta}}{1-e^{i\theta}}\right)$ as required, and $S=S'=\operatorname{Im}\left(\dfrac{1-e^{in\theta}}{1-e^{i\theta}}\right)$ (noting the leading $\sin0\theta=0$ term doesn't change the value of the sine sum, so $S'$ is the same series as $S$ asked for).
Extracting the real and imaginary parts
To split an expression like $\dfrac{1-e^{in\theta}}{1-e^{i\theta}}$ into real and imaginary parts, factor a half-angle exponential out of both the numerator and denominator, using the identity $1-e^{ik\theta}=e^{\frac{ik\theta}2}\left(e^{-\frac{ik\theta}2}-e^{\frac{ik\theta}2}\right)=-2ie^{\frac{ik\theta}2}\sin\dfrac{k\theta}2$ (which follows from $z^{-\frac12}-z^{\frac12}$-type identities you met earlier this chapter). This turns the whole expression into a single exponential times a real trig ratio, which is then easy to read off.
From Example 1, $C+iS=\dfrac{1-e^{in\theta}}{1-e^{i\theta}}$. Using $1-e^{ik\theta}=-2ie^{\frac{ik\theta}2}\sin\dfrac{k\theta}2$ on both numerator ($k=n$) and denominator ($k=1$):
$$C+iS=\frac{-2ie^{\frac{in\theta}2}\sin\frac{n\theta}2}{-2ie^{\frac{i\theta}2}\sin\frac\theta2}$$
$$=e^{i\left(\frac{n\theta}2-\frac\theta2\right)}\cdot\frac{\sin\frac{n\theta}2}{\sin\frac\theta2}$$
$$=e^{i(n-1)\frac\theta2}\cdot\frac{\sin\frac{n\theta}2}{\sin\frac\theta2}$$
Since $\dfrac{\sin\frac{n\theta}2}{\sin\frac\theta2}$ is real, this is already in the form $($real number$)\times e^{i\phi}$ with $\phi=(n-1)\dfrac\theta2$. Taking the real part:
$$C=\frac{\sin\frac{n\theta}2}{\sin\frac\theta2}\cos\left((n-1)\frac\theta2\right)$$
Using the product-to-sum identity $\sin A\cos B=\frac12\big(\sin(A+B)+\sin(A-B)\big)$ with $A=\dfrac{n\theta}2$ and $B=(n-1)\dfrac\theta2$:
$\sin\dfrac{n\theta}2\cos\left((n-1)\dfrac\theta2\right)=\dfrac12\left(\sin\left(\left(n-\dfrac12\right)\theta\right)+\sin\dfrac\theta2\right)$
So $C=\dfrac{\frac12\sin\left(\left(n-\frac12\right)\theta\right)+\frac12\sin\frac\theta2}{\sin\frac\theta2}=\dfrac{\sin\left(\left(n-\frac12\right)\theta\right)}{2\sin\frac\theta2}+\dfrac12$, as required.
Convergent infinite series with a shrinking modulus
When each term of the series carries an extra factor $r^k$ with $0\lt r\lt1$, the corresponding complex series has common ratio $z=re^{i\theta}$, and now $|z|=r<1$ — so the infinite geometric series formula $\dfrac{a}{1-z}$ applies directly, with no need to stop at $n$ terms.
$C+iS=1+\dfrac12(\cos\theta+i\sin\theta)+\dfrac14(\cos2\theta+i\sin2\theta)+\cdots=1+\dfrac12z+\dfrac14z^2+\cdots$ where $z=e^{i\theta}$, an infinite geometric series with first term $1$ and ratio $\dfrac z2$ (and $\left|\dfrac z2\right|=\dfrac12<1$, so it converges). So
$$C+iS=\frac1{1-\frac z2}=\frac2{2-z}=\frac2{2-e^{i\theta}}$$
To separate real and imaginary parts, multiply top and bottom by the conjugate of the denominator, $2-e^{-i\theta}$:
$$\frac2{2-e^{i\theta}}\cdot\frac{2-e^{-i\theta}}{2-e^{-i\theta}}=\frac{2(2-e^{-i\theta})}{(2-e^{i\theta})(2-e^{-i\theta})}=\frac{4-2e^{-i\theta}}{4-2e^{i\theta}-2e^{-i\theta}+1}$$
Using $e^{i\theta}+e^{-i\theta}=2\cos\theta$ in the denominator, and $e^{-i\theta}=\cos\theta-i\sin\theta$ in the numerator: denominator $=5-4\cos\theta$, numerator $=4-2\cos\theta+2i\sin\theta$.
So $C+iS=\dfrac{(4-2\cos\theta)+2i\sin\theta}{5-4\cos\theta}$, giving $C=\dfrac{4-2\cos\theta}{5-4\cos\theta}=\dfrac{2(2-\cos\theta)}{5-4\cos\theta}$, as required.
Before summing, always check that the series you're given for $C$ and the series implied for $S$ line up term-by-term with the same coefficients and the same angle multiples — if the question only gives you $C$, you generally need to construct the matching $S$ yourself (as in Example 1) rather than assume it's given.
Exam-style questions
Question 1 6 marks
Show that $$\cos\theta+\cos3\theta+\cos5\theta+\cdots+\cos(2n-1)\theta=\frac{\sin(2n\theta)}{2\sin\theta}$$ for $0<\theta<\pi$. (Hint: consider $C+iS$ as a geometric series with common ratio $e^{2i\theta}$.)
Let $z=e^{i\theta}$. Then $C+iS=z+z^3+z^5+\cdots+z^{2n-1}=z(1+z^2+z^4+\cdots+z^{2n-2})$, a geometric series inside the bracket with first term $1$, ratio $z^2$, and $n$ terms. (M1, A1)
$C+iS=z\cdot\dfrac{1-z^{2n}}{1-z^2}=\dfrac{z-z^{2n+1}}{1-z^2}=\dfrac{z(1-z^{2n})}{1-z^2}$. Factor half-angles: $1-z^{2n}=-2ie^{in\theta}\sin(n\theta)$ and $1-z^2=-2ie^{i\theta}\sin\theta$. (M1, A1)
$C+iS=\dfrac{e^{i\theta}\cdot(-2i)e^{in\theta}\sin(n\theta)}{-2ie^{i\theta}\sin\theta}=\dfrac{e^{in\theta}\sin(n\theta)}{\sin\theta}$. Taking the real part: $C=\dfrac{\cos(n\theta)\sin(n\theta)}{\sin\theta}=\dfrac{\frac12\sin(2n\theta)}{\sin\theta}=\dfrac{\sin(2n\theta)}{2\sin\theta}$, as required. (M1, A1)
Question 2 7 marks
The series $C$ and $S$ are defined for $0<\theta<2\pi$ by $$C=1+\frac13\cos\theta+\frac19\cos2\theta+\cdots$$ $$S=\frac13\sin\theta+\frac19\sin2\theta+\cdots$$ Show that $C+iS=\dfrac{3}{3-e^{i\theta}}$, and find $S$ in terms of $\theta$.
$C+iS=1+\dfrac13z+\dfrac19z^2+\cdots$, an infinite geometric series with $a=1$, ratio $\dfrac z3$: $C+iS=\dfrac1{1-\frac z3}=\dfrac3{3-z}=\dfrac3{3-e^{i\theta}}$. (M1, A1)
Multiply by the conjugate $3-e^{-i\theta}$: $\dfrac{3(3-e^{-i\theta})}{(3-e^{i\theta})(3-e^{-i\theta})}=\dfrac{9-3e^{-i\theta}}{9-3(e^{i\theta}+e^{-i\theta})+1}=\dfrac{9-3\cos\theta+3i\sin\theta}{10-6\cos\theta}$. (M1, A1, A1)
So $S=\dfrac{3\sin\theta}{10-6\cos\theta}$. (A1)