4.02q - Summing trigonometric series | OCR A-Level Further Mathematics A

Add trigonometric terms by putting their cosine and sine parts back into a complex exponential. You will sum finite and convergent infinite geometric series, handle their exceptional cases, and recognise binomial coefficients hidden inside trigonometric sums.

Finite geometric sums of complex exponentials

Can you find the value of

1+\frac12\cos\theta+\frac14\cos2\theta+\frac18\cos3\theta+\cdots

without adding infinitely many terms? At $\theta=\frac\pi2$ , every other term is zero, and the rest give $1-\frac14+\frac1{16}-\cdots=\frac45$ . We will find one expression that works for every real $\theta$ .

The first move is to put the sine terms back. For example,

\begin{aligned} 1+e^{i\theta}+e^{2i\theta} &=(1+\cos\theta+\cos2\theta) \\ &\quad+i(\sin\theta+\sin2\theta) \end{aligned}

The real part is the cosine sum and the imaginary part is the sine sum. The complex expression is easier to add because it is a geometric progression with ratio $e^{i\theta}$ .

You already know

Euler's formula gives $e^{ir\theta}=\cos r\theta+i\sin r\theta$ . For a geometric progression with $n$ terms, first term $a$ and ratio $q\ne1$ , the sum is $a(1-q^n)/(1-q)$ . The same cancellation proof works with complex $q$ .

Indeed, subtracting $qS$ from $S=a+aq+\cdots+aq^{n-1}$ leaves $(1-q)S=a(1-q^n)$ .

A finite complex geometric sum

For a positive integer $n$ , complex $a$ and complex $q\ne1$ ,

\sum_{r=0}^{n-1}aq^r=\frac{a(1-q^n)}{1-q}

When $q=1$ , the sum is $na$ , obtained directly rather than by dividing by zero.

In the formula booklet.

For a unit ratio it helps to rewrite a difference of exponentials symmetrically:

1-e^{iu}=e^{iu/2}(e^{-iu/2}-e^{iu/2})=-2i e^{iu/2}\sin\frac u2

This derived identity is not printed in the formula booklet. It converts the quotient into a real size times a single exponential.

Worked example 1: a finite cosine sum

Question. Determine $\displaystyle C_n=\sum_{r=0}^{n-1}\cos r\theta$ , where $\theta$ is not an integer multiple of $2\pi$ .

Plan. Sum the complex geometric progression, centre its numerator and denominator, then take the real part.

\begin{aligned} S_n&=\sum_{r=0}^{n-1}e^{ir\theta}=\frac{1-e^{in\theta}}{1-e^{i\theta}} && \text{geometric sum} \\ &=\frac{-2i e^{in\theta/2}\sin(n\theta/2)}{-2i e^{i\theta/2}\sin(\theta/2)} && \text{centre each difference} \\ &=\frac{\sin(n\theta/2)}{\sin(\theta/2)}e^{i(n-1)\theta/2} && \text{cancel factors} \end{aligned}

Why this step? The sine quotient is real, so the real part of the final expression is obtained just by replacing its exponential by its cosine part.

Answer. $\displaystyle C_n=\frac{\sin(n\theta/2)\cos((n-1)\theta/2)}{\sin(\theta/2)}$ .

Check. For $n=2$ and $\theta=\frac\pi3$ , the sum is $1+\frac12=\frac32$ . The formula gives $\frac{\sin(\pi/3)\cos(\pi/6)}{\sin(\pi/6)}=\frac32$ .

Worked example 2 changes only the part you extract: imaginary instead of real. The complex sum and all its conditions stay the same.

Worked example 2: a finite sine sum

Question. Determine $\displaystyle T_n=\sum_{r=0}^{n-1}\sin r\theta$ , where $\theta$ is not an integer multiple of $2\pi$ .

Plan. Use the same complex geometric progression and extract its imaginary part.

\begin{aligned} C_n+iT_n&=\frac{1-e^{in\theta}}{1-e^{i\theta}} && \text{combine the two sums} \\ &=\frac{\sin(n\theta/2)}{\sin(\theta/2)}e^{i(n-1)\theta/2} && \text{centre and cancel} \\ T_n&=\frac{\sin(n\theta/2)\sin((n-1)\theta/2)}{\sin(\theta/2)} && \text{imaginary part} \end{aligned}

Why this step? The imaginary part is the real coefficient of $i$ ; taking it introduces sine and leaves no $i$ in the answer.

Answer. $\displaystyle T_n=\frac{\sin(n\theta/2)\sin((n-1)\theta/2)}{\sin(\theta/2)}$ .

Check. At $n=2$ , $\theta=\frac\pi3$ , this is $\frac{\sin(\pi/3)\sin(\pi/6)}{\sin(\pi/6)}=\frac{\sqrt3}{2}$ , equal to $\sin0+\sin\frac\pi3$ .

The two real sum formulae are not in the formula booklet: obtain them from its geometric-series formula. If $\theta=2k\pi$ , each cosine term is 1 and each sine term is 0, so $C_n=n$ and $T_n=0$ . The quotient formulae then read $0/0$ and cannot be substituted into directly.

Watch out

Wrong: $\sum_{r=1}^{n}e^{ir\theta}=(1-e^{in\theta})/(1-e^{i\theta})$ . That expression starts at $r=0$ . Right: the first term is $e^{i\theta}$ , so multiply the quotient by $e^{i\theta}$ .

Changing the first index rotates the whole complex sum by one step, which changes the angle in its final cosine factor.

Convergent trigonometric sums

A factor that shrinks each term turns the complex geometric progression into a convergent infinite series.

Write $q=\rho e^{i\theta}$ . Its modulus is $|q|=|\rho|$ , because $|e^{i\theta}|=1$ . If $|\rho|<1$ , then $|q^n|=|\rho|^n\to0$ , so both the real and imaginary parts of the omitted tail tend to zero. This is the condition that permits an infinite sum.

An infinite complex geometric sum

For complex $a,q$ with $|q|<1$ ,

\sum_{r=0}^{\infty}aq^r=\frac{a}{1-q}

The modulus condition is essential; in particular, $q=e^{i\theta}$ has modulus 1 and does not give a convergent geometric series.

In the formula booklet.

The sum of a complex GP is also the sum of its real parts plus $i$ times the sum of its imaginary parts. Rationalise $1-\rho e^{i\theta}=1-\rho\cos\theta-i\rho\sin\theta$ to read them off.

Worked example 3: the opening infinite cosine sum

Question. Determine $\displaystyle C=\sum_{r=0}^{\infty}2^{-r}\cos r\theta$ .

Plan. Sum the GP with ratio $\frac12e^{i\theta}$ , then multiply by the conjugate of its denominator.

\begin{aligned} S&=\sum_{r=0}^{\infty}\left(\frac12e^{i\theta}\right)^r=\frac1{1-\frac12e^{i\theta}} && |q|=\tfrac12<1 \\ &=\frac{1-\frac12\cos\theta+\frac12i\sin\theta}{(1-\frac12\cos\theta)^2+\frac14\sin^2\theta} && \text{use the conjugate} \\ &=\frac{1-\frac12\cos\theta+\frac12i\sin\theta}{\frac54-\cos\theta} && \cos^2\theta+\sin^2\theta=1 \end{aligned}

Why this step? A real denominator makes the real and imaginary parts visible, while its squared-modulus form shows it cannot vanish here.

Answer. $\displaystyle C=\frac{4-2\cos\theta}{5-4\cos\theta}$ .

Check. At $\theta=0$ , the sum is $1+\frac12+\frac14+\cdots=2$ , and the formula gives 2. At $\theta=\frac\pi2$ , it gives $\frac45$ , answering the numerical case from the opening.

Worked example 4 changes only cosine to sine. The convergence condition and the denominator are unchanged.

Worked example 4: the matching infinite sine sum

Question. Determine $\displaystyle T=\sum_{r=0}^{\infty}2^{-r}\sin r\theta$ .

Plan. Extract the imaginary part of the same convergent complex GP after rationalising its denominator.

\begin{aligned} \sum_{r=0}^{\infty}2^{-r}e^{ir\theta} &=\frac1{1-\frac12e^{i\theta}} && |q|=\tfrac12<1 \\ &=\frac{4-2\cos\theta+2i\sin\theta}{5-4\cos\theta} && \text{rationalise; multiply by }4 \\ T&=\frac{2\sin\theta}{5-4\cos\theta} && \text{imaginary part} \end{aligned}

Why this step? The complex sum solves the cosine and sine problems together; extracting a different part requires no new summation.

Answer. $\displaystyle T=\frac{2\sin\theta}{5-4\cos\theta}$ .

Check. At $\theta=\frac\pi2$ , the series is $\frac12-\frac18+\frac1{32}-\cdots=\frac{1/2}{1+1/4}=\frac25$ , agreeing with the formula.

These rational trigonometric sum expressions are derived results, not printed in the formula booklet. More generally the denominator is $1-2\rho\cos\theta+\rho^2$ , the squared modulus of $1-\rho e^{i\theta}$ .

Watch out

Wrong: an infinite sum of $e^{ir\theta}$ equals $1/(1-e^{i\theta})$ whenever the denominator is non-zero. A non-zero denominator is not a convergence test. Right: use the infinite GP formula only when the modulus of the ratio is less than 1.

Changing the shrinking factor changes the denominator's squared modulus; the extraction of the imaginary part stays the same.

Binomial coefficients in trigonometric sums

When the coefficients are binomial coefficients, use a binomial expansion instead of a geometric sum.

For example, $1,4,6,4,1$ are the coefficients of $(1+z)^4$ , so

1+4e^{i\theta}+6e^{2i\theta}+4e^{3i\theta}+e^{4i\theta}=(1+e^{i\theta})^4

The coefficients change, so this is not a GP. Its real and imaginary parts are still the desired trigonometric sums.

The binomial sum

For a non-negative integer $n$ and complex $z$ ,

\sum_{r=0}^{n}\binom nr z^r=(1+z)^n

In the formula booklet.

Here the sum has $n+1$ terms. To simplify the right side, centre the exponential sum:

1+e^{i\theta}=e^{i\theta/2}(e^{-i\theta/2}+e^{i\theta/2})=2\cos\frac\theta2\ e^{i\theta/2}

This derived identity is not separately printed. It is valid even when $\cos(\theta/2)$ is negative: the expression is an algebraic product, and we are not claiming its real prefactor is a modulus.

Worked example 5: a binomial cosine sum

Question. Determine $\displaystyle\sum_{r=0}^{n}\binom nr\cos r\theta$ , for a positive integer $n$ .

Plan. Recognise the real part of $(1+e^{i\theta})^n$ , centre the bracket, and use de Moivre.

\begin{aligned} S&=\sum_{r=0}^{n}\binom nr e^{ir\theta}=(1+e^{i\theta})^n && \text{binomial theorem} \\ &=\left(2\cos\frac\theta2\right)^n e^{in\theta/2} && \text{centre; take the integer power} \\ \mathrm{Re}(S)&=2^n\cos^n\frac\theta2\cos\frac{n\theta}2 && \text{real part} \end{aligned}

Why this step? The entire coefficient $(2\cos(\theta/2))^n$ is real, so it stays outside when you extract the real part.

Answer. $\displaystyle\sum_{r=0}^{n}\binom nr\cos r\theta=2^n\cos^n\frac\theta2\cos\frac{n\theta}2$ .

Check. For $n=2$ , $\theta=\frac\pi3$ , the sum is $1+2(\frac12)-\frac12=\frac32$ . The expression is $4(\frac{\sqrt3}{2})^2(\frac12)=\frac32$ .

Worked example 6 changes only the part extracted, so the sine sum follows from the very same expansion.

Worked example 6: a binomial sine sum

Question. Determine $\displaystyle\sum_{r=0}^{n}\binom nr\sin r\theta$ , for a positive integer $n$ .

Plan. Sum the complex binomial expression and extract its imaginary part.

\begin{aligned} S&=(1+e^{i\theta})^n && \text{binomial theorem} \\ &=2^n\cos^n\frac\theta2\left(\cos\frac{n\theta}2+i\sin\frac{n\theta}2\right) && \text{de Moivre} \\ \mathrm{Im}(S)&=2^n\cos^n\frac\theta2\sin\frac{n\theta}2 && \text{coefficient of }i \end{aligned}

Why this step? Taking imaginary parts gives the sine sum directly, including its sign; no division by a trigonometric expression is needed.

Answer. $\displaystyle\sum_{r=0}^{n}\binom nr\sin r\theta=2^n\cos^n\frac\theta2\sin\frac{n\theta}2$ .

Check. At $n=2$ , $\theta=\frac\pi3$ , the sum is $2(\frac{\sqrt3}{2})+\frac{\sqrt3}{2}=\frac{3\sqrt3}{2}$ , also the value of the formula.

The binomial cosine and sine sum results are not printed in the formula booklet: they come from its binomial theorem and de Moivre's theorem. A factor $a^r$ changes the bracket to $1+ae^{i\theta}$ ; alternating signs change it to $1-e^{i\theta}$ . Recognising the coefficients tells you which expansion to choose.

Problem-solving

Combine cosine and sine first, then inspect the coefficients: constant ratio suggests a GP; binomial coefficients suggest a power of a bracket.

An angle step of $2\theta$ simply replaces $e^{i\theta}$ by $e^{2i\theta}$ inside the binomial bracket.

Quick Recap

Add $e^{ir\theta}$ terms first; their real and imaginary parts give cosine and sine sums.
Count the terms and identify the first term before using a finite GP formula.
Handle ratio 1 directly; do not substitute into a zero denominator.
An infinite complex GP requires its ratio to have modulus less than 1.
Binomial coefficients call for $(1+z)^n$ , followed by de Moivre and extraction of a real or imaginary part.

The opening infinite series is $(4-2\cos\theta)/(5-4\cos\theta)$ , so at $\theta=\frac\pi2$ it is exactly $\frac45$ .

Last Accessed

Pure Core

Statistics

Mechanics

Discrete Mathematics

Additional Pure Mathematics

4.02q - Summing trigonometric series

Finite geometric sums of complex exponentials

Convergent trigonometric sums

Binomial coefficients in trigonometric sums

Quick Recap

Learn faster with Nova