Pj13 - Summing trigonometric series | OCR A-Level Further Mathematics B (MEI)

Add a sine series and a cosine series together as one complex series, sum it, then extract the part you need. This second use of Pj13 covers binomially weighted sums, finite geometric sums and convergent infinite series, including series whose angles advance by more than one step.

Binomial coefficients reveal a power

What is the exact value of

1+8\cos\frac\pi3+28\cos\frac{2\pi}3+56\cos\pi+70\cos\frac{4\pi}3+56\cos\frac{5\pi}3+28\cos2\pi+8\cos\frac{7\pi}3+\cos\frac{8\pi}3?

The coefficients look far less mysterious when you recognise row $8$ of Pascal's triangle. One binomial expansion can replace the whole sum.

You already know

For $z=\cos\theta+i\sin\theta$ , de Moivre gives $z^r=\cos r\theta+i\sin r\theta$ . The binomial theorem gives $(1+z)^n=\sum_{r=0}^{n}\binom nr z^r$ for a non-negative integer $n$ .

Consequently, the real and imaginary parts of $(1+z)^n$ are the two weighted sums. The half-angle identities give a useful factorisation:

\begin{aligned} 1+z&=1+\cos\theta+i\sin\theta \\ &=2\cos\frac\theta2\left(\cos\frac\theta2+i\sin\frac\theta2\right) \end{aligned}

This is an algebraic identity even when $\cos(\theta/2)$ is negative; do not call that negative coefficient a modulus. Raising the factorisation to an integer power and using de Moivre now sums both series at once.

Binomially weighted trigonometric sums

For a positive integer $n$ and real $\theta$ ,

\begin{aligned} \sum_{r=0}^{n}\binom nr\cos r\theta &=2^n\cos^n\frac\theta2\cos\frac{n\theta}2 \\ \sum_{r=0}^{n}\binom nr\sin r\theta &=2^n\cos^n\frac\theta2\sin\frac{n\theta}2 \end{aligned}

Not in the formula booklet: learn it.

The booklet supplies the binomial theorem and de Moivre's theorem separately. In an exam, explain how you combine them rather than simply quoting a remembered sum.

Worked example 1: extracting a cosine sum

Question. Find $\sum_{r=0}^{4}\binom4r\cos(r\pi/3)$ exactly.

Plan. Make the sum the real part of $(1+z)^4$ , where $z=\cos(\pi/3)+i\sin(\pi/3)$ .

\begin{aligned} 1+z&=\sqrt3\left(\cos\frac\pi6+i\sin\frac\pi6\right) && \text{half-angle factorisation} \\ (1+z)^4&=9\left(\cos\frac{2\pi}3+i\sin\frac{2\pi}3\right) && \text{de Moivre} \\ \sum_{r=0}^{4}\binom4r\cos\frac{r\pi}3&=9\cos\frac{2\pi}3=-\frac92 && \text{real parts} \end{aligned}

Why this step? The binomial coefficient multiplies the whole power $z^r$ , so its real part is exactly the coefficient times $\cos(r\pi/3)$ .

Answer. $-9/2$ .

Check. Directly, $1+4(1/2)+6(-1/2)+4(-1)+(-1/2)=-9/2$ .

Change only the requested part of the same complex sum. There is no need to repeat four separate trigonometric calculations.

Worked example 2: extracting a sine sum

Question. Find $\sum_{r=0}^{4}\binom4r\sin(r\pi/3)$ exactly.

Plan. Use the imaginary part of the same binomial expansion.

\begin{aligned} \sum_{r=0}^{4}\binom4r z^r&=(1+z)^4 && \text{binomial theorem} \\ &=9\left(-\frac12+\frac{\sqrt3}{2}i\right) && \text{de Moivre} \\ \sum_{r=0}^{4}\binom4r\sin\frac{r\pi}3&=\frac{9\sqrt3}{2} && \text{imaginary parts} \end{aligned}

Why this step? The imaginary part is the coefficient of $i$ , so the real sum you seek does not contain $i$ .

Answer. $9\sqrt3/2$ .

Check. The non-zero sine terms give $4(\sqrt3/2)+6(\sqrt3/2)-\sqrt3/2=9\sqrt3/2$ .

The surprising feature is that an entire row of Pascal's triangle, with alternating positive and negative trigonometric contributions, collapses to one point on a circle. For the opening sum that point has modulus $81$ and angle $4\pi/3$ , so its real part will be $-81/2$ .

Watch out

Wrong: $\sum\binom nr\cos r\theta=(1+\cos\theta)^n$ .

Right: take $\operatorname{Re}((1+z)^n)$ . Taking real parts before raising to a power loses all the terms involving even powers of $i$ .

A new exponent tests whether you can recognise the structure without expanding the entire row.

Finite sums with equally spaced angles

Without binomial coefficients, equally spaced angles usually signal a geometric series. The common ratio is complex, but the cancellation proof of the geometric-series formula still works.

If $C=1+\cos\theta+\cdots+\cos n\theta$ and $S=\sin\theta+\cdots+\sin n\theta$ , then $C+iS=1+z+\cdots+z^n$ , where $z=\cos\theta+i\sin\theta$ .

Finite geometric sum

For complex $q\ne1$ and positive integer $N$ ,

1+q+\cdots+q^{N-1}=\frac{1-q^N}{1-q}

When $q=1$ , the sum is $N$ , so do not use the fraction.

In the formula booklet.

To extract real and imaginary parts, either rationalise the denominator or factor numerator and denominator into half-angle forms. The latter often keeps a general finite sum pleasantly short.

Worked example 3: consecutive angles

Question. Show that, for $\theta$ not a multiple of $2\pi$ ,

\sum_{r=0}^{n}\cos r\theta =\frac{\sin((n+1)\theta/2)\cos(n\theta/2)}{\sin(\theta/2)}

Plan. Sum the complex geometric series and use half-angle factorisations before taking real parts.

Write $u_\alpha=\cos\alpha+i\sin\alpha$ . The identity
$1-u_\alpha=-2i\sin(\alpha/2)u_{\alpha/2}$ follows by expanding the right side. Therefore

\begin{aligned} C+iS&=\frac{1-z^{n+1}}{1-z} && \text{geometric sum} \\ &=\frac{-2i\sin((n+1)\theta/2)u_{(n+1)\theta/2}} {-2i\sin(\theta/2)u_{\theta/2}} && \text{factor both differences} \\ &=\frac{\sin((n+1)\theta/2)}{\sin(\theta/2)}u_{n\theta/2} && \text{divide unit complex numbers} \end{aligned}

Why this step? Dividing the unit factors subtracts their angles. The remaining prefactor is real, so its real part is obtained by multiplying by $\cos(n\theta/2)$ .

Answer. Taking real parts gives the stated result. Taking imaginary parts also gives

\sum_{r=0}^{n}\sin r\theta =\frac{\sin((n+1)\theta/2)\sin(n\theta/2)}{\sin(\theta/2)}

Check. At $n=1$ , the cosine formula gives $2\cos^2(\theta/2)=1+\cos\theta$ .

Now change only the angle step, from $\theta$ to $2\theta$ . The ratio must change with it.

Worked example 4: an angle step of twice theta

Question. Find an expression for $1+\cos2\theta+\cos4\theta+\cdots+\cos2n\theta$ , when $\theta$ is not a multiple of $\pi$ .

Plan. Sum $1+z^2+z^4+\cdots+z^{2n}$ , whose common ratio is $z^2$ .

\begin{aligned} C+iS&=\frac{1-z^{2(n+1)}}{1-z^2} && \text{ratio }z^2 \\ &=\frac{-2i\sin((n+1)\theta)u_{(n+1)\theta}} {-2i\sin\theta\,u_\theta} && \text{half-angle factors} \\ &=\frac{\sin((n+1)\theta)}{\sin\theta}u_{n\theta} && \text{subtract arguments} \end{aligned}

Why this step? There are still $n+1$ terms. The final exponent is $2n$ , but the geometric numerator uses the ratio to the power $n+1$ .

Answer. $C=\dfrac{\sin((n+1)\theta)\cos n\theta}{\sin\theta}$ .

Check. At $n=2$ , $\theta=\pi/3$ , the direct sum is $1-1/2-1/2=0$ , and the formula has numerator $\sin\pi=0$ .

The two extracted finite trigonometric-sum identities are not in the formula booklet: learn the derivation. At the excluded angles, return to the original terms. For example, when $\theta=2k\pi$ , the consecutive-angle cosine sum is $n+1$ , although its fractional expression gives $0/0$ .

Watch out

Wrong: $1+z+\cdots+z^n=(1-z^n)/(1-z)$ .

Right: the numerator is $1-z^{n+1}$ , because the sum contains $n+1$ terms. Count terms before applying the formula.

Use the imaginary-part formula for this exact-value check; its first term is $\sin0=0$ .

Infinite sums and shifted angles

An infinite complex geometric series converges when its ratio has modulus less than $1$ . This is a distance condition: repeated powers approach the origin, whatever their arguments do.

For example, $q=\frac12(\cos\theta+i\sin\theta)$ has $|q|=1/2$ , so each power has half the previous modulus. Its cosine and sine parts can oscillate while the series still converges.

Infinite complex geometric sum

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

a+aq+aq^2+\cdots=\frac{a}{1-q}

The finite formula tends to this result because $q^N\to0$ .

In the formula booklet.

After summing, make the denominator real using its conjugate. Taking a real or imaginary part of a quotient before doing this is generally invalid.

Worked example 5: a convergent sine series

Question. Show that

\frac12\sin\theta+\frac14\sin2\theta+\frac18\sin3\theta+\cdots =\frac{2\sin\theta}{5-4\cos\theta}

Plan. Sum $q+q^2+\cdots$ , where $q=z/2$ , rationalise, then take imaginary parts.

\begin{aligned} C+iS&=\frac{z/2}{1-z/2}=\frac{z}{2-z} && |z/2|<1 \\ &=\frac{z(2-z^{-1})}{(2-z)(2-z^{-1})} && z^*=z^{-1} \\ &=\frac{2z-1}{5-2(z+z^{-1})} && zz^{-1}=1 \\ &=\frac{(2\cos\theta-1)+2i\sin\theta}{5-4\cos\theta} && \text{real denominator} \end{aligned}

Why this step? Both numerator and denominator must be multiplied by the denominator's conjugate. The denominator is then $|2-z|^2$ , a positive real number.

Answer. Taking imaginary parts gives $S=\dfrac{2\sin\theta}{5-4\cos\theta}$ , as required.

Check. At $\theta=\pi/2$ , the series is $1/2-1/8+1/32-\cdots=(1/2)/(1+1/4)=2/5$ , matching the formula.

Change the angle step from $\theta$ to $2\theta$ , keeping the initial angle $\theta$ . The first term and the common ratio now have different arguments.

Worked example 6: a different initial angle and step

Question. Sum

C=\cos\theta+\frac12\cos3\theta+\frac14\cos5\theta+\cdots

Plan. Pair this with the corresponding sine sum and use first term $z$ , ratio $z^2/2$ .

\begin{aligned} C+iS&=\frac{z}{1-z^2/2}=\frac{2z}{2-z^2} && |z^2/2|=1/2 \\ &=\frac{2z(2-z^{-2})}{(2-z^2)(2-z^{-2})} && \text{conjugate denominator} \\ &=\frac{4z-2z^{-1}}{5-2(z^2+z^{-2})} && \text{simplify powers} \\ &=\frac{2\cos\theta+6i\sin\theta}{5-4\cos2\theta} && \text{de Moivre} \end{aligned}

Why this step? The ratio comes from the change between successive terms: the coefficient halves and the angle increases by $2\theta$ . It is not simply half the first term.

Answer. $C=\dfrac{2\cos\theta}{5-4\cos2\theta}$ .

Check. At $\theta=0$ , the series is $1+1/2+1/4+\cdots=2$ , and the expression is $2/(5-4)=2$ .

The derived trigonometric sums are not in the formula booklet: learn the complex-series method. Notice that $5-4\cos2\theta\ge1$ , so the final denominator cannot vanish for a real angle. That agrees with convergence for every real $\theta$ .

Exam tip

State the first term, common ratio and its modulus before summing an infinite series. When establishing a given sine or cosine result, show a real denominator before extracting the required part.

Finally, the opening binomial sum is the real part of $(1+z)^8$ at $\theta=\pi/3$ : $(\sqrt3)^8\cos(4\pi/3)=-81/2$ . The long string of cosines hid a single binomial power.

Watch out

Wrong: the ratio in $z+\frac12z^3+\frac14z^5+\cdots$ is $z/2$ . Each exponent increases by $2$ , so dividing the second term by the first gives the correct ratio $z^2/2$ .

Find the next series’ ratio from two consecutive complex terms.

Quick Recap

Binomial coefficients suggest $(1+z)^n$ ; its real and imaginary parts give weighted cosine and sine sums.
Equally spaced angles suggest a geometric series. Identify the angle step and count its terms.
For infinite sums, check that the modulus of the common ratio is below $1$ .
Make a complex denominator real before extracting parts; a shifted first angle changes the first term, while the angle step changes the ratio.

The opening nine-term cosine sum is $-81/2$ , obtained from the real part of one eighth power.

Last Accessed

Core Pure

Mechanics (Major and Minor)

Statistics (Major and Minor)

Modelling with Algorithms

Numerical Methods

Extra Pure

Further Pure with Technology

Pj13 - Summing trigonometric series

Binomial coefficients reveal a power

Finite sums with equally spaced angles

Infinite sums and shifted angles

Quick Recap

Learn faster with Nova