2024 H2 Maths Paper 1 Q7 — A Telescoping Trigonometric Sum

The chain, livethe cancellation

How many terms,

n

The angle

\theta

0.42 rad

Each difference

f(2r-1)-f(2r+1)

The two survivors Cancelled in pairs

Part (a)

2 marks

Given $f(r)=\cos(r\theta)$ , show that $f(2r-1)-f(2r+1)=2\sin\theta\sin(2r\theta)$ .

Approach

A difference of two cosines on the left, a product of two sines on the right — that is the factor formula $\cos A-\cos B=-2\sin\frac{A+B}{2}\sin\frac{A-B}{2}$ , straight from MF26. Identify $A$ and $B$ , compute the half-sum and half-difference, and the rest is a sign.

Writing out the left-hand side:

f(2r-1)-f(2r+1)=\cos\left((2r-1)\theta\right)-\cos\left((2r+1)\theta\right)

With $A=(2r-1)\theta$ and $B=(2r+1)\theta$ :

\frac{A+B}{2}=\frac{(2r-1)\theta+(2r+1)\theta}{2}=\frac{4r\theta}{2}=2r\theta,\qquad \frac{A-B}{2}=\frac{-2\theta}{2}=-\theta

\begin{aligned}\cos A-\cos B&=-2\sin(2r\theta)\sin(-\theta)\\&=-2\sin(2r\theta)\left(-\sin\theta\right)\\&=2\sin\theta\sin(2r\theta)\quad\textbf{(shown)}\end{aligned}

To check your answer:

Take $r=1$ , $\theta=\frac{\pi}{6}$ : the left side is $\cos\frac{\pi}{6}-\cos\frac{\pi}{2}=\frac{\sqrt{3}}{2}-0=\frac{\sqrt{3}}{2}$ , and the right side is $2\left(\frac{1}{2}\right)\sin\frac{\pi}{3}=\frac{\sqrt{3}}{2}$ ✓.

Where the marks go

The sign. $\sin(-\theta)=-\sin\theta$ , and that minus is what cancels the one in front of the factor formula. Losing it gives $-2\sin\theta\sin(2r\theta)$ , which then makes part (b) come out negative — and since (b) is also a show, the error surfaces two marks later with nowhere to hide.

Part (b)

3 marks

Hence, given $\sin\theta\neq0$ , show that $\displaystyle\sum_{r=1}^{n}\sin(2r\theta)=\frac{\cos\theta-\cos\left((2n+1)\theta\right)}{2\sin\theta}$ .

Approach

Part (a) has written the thing you want to sum as a difference of consecutive values of the same function. Summing such a thing is the method of differences: write out the first few and the last few, and everything in the middle cancels in pairs.

Part (a) can be read from right to left:

2\sin\theta\sin(2r\theta)=\cos\left((2r-1)\theta\right)-\cos\left((2r+1)\theta\right)

Summing both sides from $r=1$ to $n$ :

2\sin\theta\sum_{r=1}^{n}\sin(2r\theta)=\sum_{r=1}^{n}\left[\cos\left((2r-1)\theta\right)-\cos\left((2r+1)\theta\right)\right]

Writing the right-hand side out term by term:

\begin{aligned}&\left[\cos\theta-\cos3\theta\right]\\+\ &\left[\cos3\theta-\cos5\theta\right]\\+\ &\left[\cos5\theta-\cos7\theta\right]\\ &\ \ \vdots\\+\ &\left[\cos\left((2n-1)\theta\right)-\cos\left((2n+1)\theta\right)\right]\end{aligned}

Each negative term is cancelled by the positive term on the next line, so only the very first and the very last survive:

2\sin\theta\sum_{r=1}^{n}\sin(2r\theta)=\cos\theta-\cos\left((2n+1)\theta\right)

Since $\sin\theta\neq0$ we may divide by $2\sin\theta$ :

\sum_{r=1}^{n}\sin(2r\theta)=\frac{\cos\theta-\cos\left((2n+1)\theta\right)}{2\sin\theta}\quad\textbf{(shown)}

To check your answer:

Take $n=1$ . The left side is $\sin2\theta$ ; the right side is $\frac{\cos\theta-\cos3\theta}{2\sin\theta}$ , which by part (a) with $r=1$ is $\frac{2\sin\theta\sin2\theta}{2\sin\theta}=\sin2\theta$ ✓. Lengthen the chain in the model and watch the intermediate values pair off.

Note — why the condition $\sin\theta\neq0$ is stated

It is not a technicality about dividing by zero for its own sake. If $\sin\theta=0$ then every term $\sin(2r\theta)$ is zero, the sum is zero, and the right-hand side is $\frac{0}{0}$ — the formula genuinely fails to say anything. The question supplies the condition because the identity is false without it.

Part (c)

3 marks

Hence find the three possible values of $\displaystyle\sum_{r=1}^{n}\sin\left(\frac{\pi r}{6}\right)\cos\left(\frac{\pi r}{6}\right)$ .

Approach

"Hence" again: get the summand into the shape $\sin(2r\theta)$ . The double-angle identity $\sin x\cos x=\frac{1}{2}\sin2x$ does it in one step, and reading off $\theta$ is then automatic. The "three possible values" at the end is the real question: it is asking what $\cos\left((2n+1)\theta\right)$ can be as $n$ runs over the positive integers.

Using $\sin x\cos x=\frac{1}{2}\sin2x$ with $x=\frac{\pi r}{6}$ :

\sin\left(\frac{\pi r}{6}\right)\cos\left(\frac{\pi r}{6}\right)=\frac{1}{2}\sin\left(\frac{\pi r}{3}\right)=\frac{1}{2}\sin\left(2r\cdot\frac{\pi}{6}\right)

so this is part (b) with $\theta=\frac{\pi}{6}$ , halved. Since $\sin\frac{\pi}{6}=\frac{1}{2}\neq0$ , the formula applies:

\begin{aligned}\sum_{r=1}^{n}\sin\left(\frac{\pi r}{6}\right)\cos\left(\frac{\pi r}{6}\right)&=\frac{1}{2}\cdot\frac{\cos\frac{\pi}{6}-\cos\left(\frac{(2n+1)\pi}{6}\right)}{2\sin\frac{\pi}{6}}\\[3pt]&=\frac{1}{2}\cdot\frac{\frac{\sqrt{3}}{2}-\cos\left(\frac{(2n+1)\pi}{6}\right)}{1}\\[3pt]&=\frac{\sqrt{3}}{4}-\frac{1}{2}\cos\left(\frac{(2n+1)\pi}{6}\right)\end{aligned}

Now the only thing that varies is $\cos\left(\frac{(2n+1)\pi}{6}\right)$ . As $n$ increases, $2n+1$ runs through the odd numbers, and $\frac{(2n+1)\pi}{6}$ steps round the circle in jumps of $\frac{\pi}{3}$ , repeating after six steps. The six angles visited are

\frac{\pi}{6},\ \frac{3\pi}{6},\ \frac{5\pi}{6},\ \frac{7\pi}{6},\ \frac{9\pi}{6},\ \frac{11\pi}{6}

and their cosines take only three distinct values, since the angles pair off symmetrically about the horizontal axis:

\cos\left(\frac{(2n+1)\pi}{6}\right)\in\left\{\frac{\sqrt{3}}{2},\ 0,\ -\frac{\sqrt{3}}{2}\right\}

Substituting each in turn:

\begin{aligned}\cos(\cdots)=\tfrac{\sqrt{3}}{2}&\ \Longrightarrow\ \text{sum}=\tfrac{\sqrt{3}}{4}-\tfrac{\sqrt{3}}{4}=0\\[2pt]\cos(\cdots)=0&\ \Longrightarrow\ \text{sum}=\tfrac{\sqrt{3}}{4}\\[2pt]\cos(\cdots)=-\tfrac{\sqrt{3}}{2}&\ \Longrightarrow\ \text{sum}=\tfrac{\sqrt{3}}{4}+\tfrac{\sqrt{3}}{4}=\tfrac{\sqrt{3}}{2}\end{aligned}

To check your answer:

Add the terms directly for the first few $n$ . At $n=1$ : $\sin\frac{\pi}{6}\cos\frac{\pi}{6}=\frac{1}{2}\cdot\frac{\sqrt{3}}{2}=\frac{\sqrt{3}}{4}$ ✓. At $n=2$ a further $\sin\frac{\pi}{3}\cos\frac{\pi}{3}=\frac{\sqrt{3}}{4}$ brings it to $\frac{\sqrt{3}}{2}$ ✓. At $n=5$ and $n=6$ the running total is back to 0 ✓. Step $n$ through the model and the total cycles through the same three numbers forever.

$\therefore$ the three possible values are $0$ , $\dfrac{\sqrt{3}}{4}$ and $\dfrac{\sqrt{3}}{2}$ .

Where the marks go

Stopping at $\frac{\sqrt{3}}{4}-\frac{1}{2}\cos\left(\frac{(2n+1)\pi}{6}\right)$ answers a question that was not asked. The words "three possible values" are an instruction to finish the job: say why the angle repeats, list what the cosine can be, and give three numbers. The other slip is forgetting the $\frac{1}{2}$ from the double-angle step, which triples every answer.

What this question was really testing

Reading a factor formula backwards. Part (a) gives you the target; the work is matching $A$ and $B$ and surviving the sign of $\sin(-\theta)$ .
The method of differences. Once a summand is written as $f(k)-f(k+2)$ , summing it is bookkeeping. Writing out three terms at the start and two at the end is what earns the mark, not asserting "it telescopes".
Forcing a sum into a known shape. $\sin x\cos x=\frac{1}{2}\sin2x$ is what makes part (c) a case of part (b) rather than a new problem.
Periodicity as the last step. A trigonometric expression in $n$ takes finitely many values when the angle steps by a rational multiple of $\pi$ . Counting them is the actual question.

Every term cancels its neighbour, and only the ends survive

Part (a)

Part (b)

Part (c)

What this question was really testing

This is how we mark, and how we teach.