Specialist Mathematics · Past QCAA questions All complex familiar questions
2020 · Paper 1 · Complex familiar

Question 16

[6 marks] Technology-free
Topic guidePremiumUnit 3 Topic 2 · Summation proofs
Game planPremium2.1 Proving a sum by induction
The question

Given $\cos(\theta)\neq0\ \forall\ n\in Z^+$, use mathematical induction to prove

[6 marks]

$$\begin{aligned}&\cos(\theta)-\cos(3\theta)+\cos(5\theta)-\ldots\\&\quad+(-1)^{n+1}\cos\big((2n-1)\theta\big)\\&=\frac{1-(-1)^n\cos(2n\theta)}{2\cos(\theta)}\end{aligned}$$

Watch the situation first

Multiply by the denominator and the sum collapses

Each term is multiplied by $2\cos\theta$. Watch the orange lines. The pieces cancel in pairs down the staircase.

Multiply each term by 2 cos θ and it splits in two Each new row cancels the piece left by the row before Only the first and last pieces survive Divide by 2 cos θ to get the right-hand side 2 cos A cos B = cos(A − B) + cos(A + B) 2 cos θ × cos θ = cos 0 + cos 2θ 2 cos θ × (−cos 3θ) = − cos 2θ − cos 4θ 2 cos θ × cos 5θ = + cos 4θ + cos 6θ 2 cos θ × (−cos 7θ) = − cos 6θ − cos 8θ Add the rows: 1 − cos 8θ 2 cos θ × (cos θ − cos 3θ + cos 5θ − cos 7θ) = 1 − cos 8θ For any n, 2 cos θ × (sum of n terms) = 1 − (−1)n cos 2nθ
Multiply each term by 2 cos θand it splits in two Each new row cancels the pieceleft by the row before Only the first and last piecessurvive Divide by 2 cos θ to getthe right-hand side 2 cos A cos B = cos(A − B) + cos(A + B) 2 cos θ × cos θ = cos 0 + cos 2θ 2 cos θ × (−cos 3θ) = − cos 2θ − cos 4θ 2 cos θ × cos 5θ = + cos 4θ + cos 6θ 2 cos θ × (−cos 7θ) = − cos 6θ − cos 8θ Add the rows: 1 − cos 8θ 2 cos θ × (cos θ − cos 3θ + cos 5θ − cos 7θ)= 1 − cos 8θ For any n, 2 cos θ × (sum of n terms)= 1 − (−1)n cos 2nθ

The induction proof does this one row at a time. When you add the next term, its first piece cancels the last piece of the formula for $n=k$, and a new last piece appears with the opposite sign.

Now for the mathematics

Add the next term and land on the next formula

Choose $\theta$ and $k$. The grey dots are the sums of the first $n$ terms, and the blue rings are the right-hand side for each $n$. The orange arrow adds the next term to the sum for $n=k$.

n

Grey dots show the sum of the first $n$ terms. Blue rings show the right-hand side. The black dot is the sum for $n=k$.

The inductive step, checked with these numbers

Every blue ring sits on its grey dot, so the formula matches the sum. The arrow adds the next term to the formula for $n=k$ and lands exactly on the formula for $n=k+1$. Testing values only checks these cases. The inductive step shows this landing works for every $k$.

For this angle the next term is 0, so the sum stays where it is. The formula for $n=k+1$ gives the same value.

Here $\cos\theta=0$, so the right-hand side divides by zero and has no value, even though every term on the left is 0. This is why the question begins with $\cos(\theta)\neq0$.

QCAA marking guide · 6 marks

Work through the solution one mark at a time

Try each step yourself before you reveal it. This follows the QCAA’s sample response, which works from the left-hand side in the inductive step.

Step 1 · initial statement1 mark

Let $n=1$. The left-hand side is just the first term. On the right, $(-1)^1=-1$, and the double angle identity $\cos(2\theta)=2\cos^2(\theta)-1$ finishes it.

$$\begin{aligned}\text{LHS}&=\cos(\theta)\\[4pt]\text{RHS}&=\frac{1-(-1)\cos(2\theta)}{2\cos(\theta)}\\[4pt]&=\frac{1+\cos(2\theta)}{2\cos(\theta)}\\[4pt]&=\frac{1+2\cos^2(\theta)-1}{2\cos(\theta)}\\[4pt]&=\cos(\theta)=\text{LHS}\end{aligned}$$

So the statement is true for $n=1$. Naming the statement $P(n)$ and writing LHS and RHS are not required, and an equivalent setting out of the initial statement is accepted.

Marker: correctly proves the initial statement

Step 2 · assumption and target1 mark

Assume the statement is true for $n=k$, where $k\in Z^+$.

$$\begin{aligned}&\cos(\theta)-\cos(3\theta)+\cos(5\theta)-\ldots\\&\quad+(-1)^{k+1}\cos\big((2k-1)\theta\big)\\&=\frac{1-(-1)^k\cos(2k\theta)}{2\cos(\theta)}\end{aligned}$$

Then write what you need to prove for $n=k+1$. Putting $n=k+1$ into the general term $(-1)^{n+1}\cos\big((2n-1)\theta\big)$ gives the extra term $(-1)^{k+2}\cos\big((2k+1)\theta\big)$.

$$\begin{aligned}&\cos(\theta)-\cos(3\theta)+\cos(5\theta)-\ldots\\&\quad+(-1)^{k+1}\cos\big((2k-1)\theta\big)\\&\quad+(-1)^{k+2}\cos\big((2k+1)\theta\big)\\&=\frac{1-(-1)^{k+1}\cos\big(2(k+1)\theta\big)}{2\cos(\theta)}\end{aligned}$$

Marker: correctly states the assumption and the proof requirement for the inductive step

Step 3 · inductive step1 mark

Start from the left-hand side for $n=k+1$. Everything up to the $k$th term is the left side of your assumption, so replace it with the right side of the assumption. Say in your working that this line uses the assumption.

$$\begin{aligned}\text{LHS}&=\frac{1-(-1)^k\cos(2k\theta)}{2\cos(\theta)}\\&\quad+(-1)^{k+2}\cos\big((2k+1)\theta\big)\end{aligned}$$

Any equivalent expression that substitutes your stated assumption is accepted. If an earlier slip carries through to here, you can still earn this mark.

Marker: uses assumption in the proof of the inductive step

Step 4 · inductive step1 mark

Put everything over the common denominator $2\cos(\theta)$. The numerator now holds a product of two cosines.

$$\text{LHS}=\frac{\begin{aligned}&1-(-1)^k\cos(2k\theta)\\&{}+(-1)^{k+2}\,2\cos(\theta)\cos\big((2k+1)\theta\big)\end{aligned}}{2\cos(\theta)}$$

Adding the angle sum and difference identities for cosine turns a product into a sum.

$$\begin{aligned}&\cos(A-B)+\cos(A+B)\\&=2\cos A\cos B\end{aligned}$$

With $A=(2k+1)\theta$ and $B=\theta$, this gives

$$\begin{aligned}&2\cos(\theta)\cos\big((2k+1)\theta\big)\\&=\cos(2k\theta)+\cos\big((2k+2)\theta\big)\end{aligned}$$

$$\text{LHS}=\frac{\begin{aligned}&1-(-1)^k\cos(2k\theta)\\&{}+(-1)^{k+2}\cos(2k\theta)\\&{}+(-1)^{k+2}\cos\big((2k+2)\theta\big)\end{aligned}}{2\cos(\theta)}$$

An equivalent expression is accepted, for example one that already uses $(-1)^k=(-1)^{k+2}$.

Marker: determines a simplified expression based on the use of a common denominator and a suitable trigonometric product identity

Step 5 · inductive step1 mark

The two $\cos(2k\theta)$ terms cancel, because the powers of $-1$ are equal. This is the cancellation from the animation.

$$(-1)^{k+2}=(-1)^k\times(-1)^2=(-1)^k$$

$$\text{LHS}=\frac{1+(-1)^{k+2}\cos\big((2k+2)\theta\big)}{2\cos(\theta)}$$

Equivalent forms are accepted, such as this one.

$$\frac{1+(-1)^{k+2}\cos\big(2(k+1)\theta\big)}{2\cos(\theta)}$$

Marker: determines a simplified expression based on the recognition that $(-1)^k=(-1)^{k+2}$

Step 6 · conclusion1 mark

Match the target. Split off one factor of $-1$ so the power matches the right-hand side for $n=k+1$. This turns the plus into a minus.

$$\begin{aligned}(-1)^{k+2}&=(-1)^{k+1}(-1)^1\\&=-(-1)^{k+1}\end{aligned}$$

$$\begin{aligned}\text{LHS}&=\frac{1-(-1)^{k+1}\cos\big(2(k+1)\theta\big)}{2\cos(\theta)}\\[4pt]&=\text{RHS}\end{aligned}$$

So the statement is true for $n=k+1$ whenever it is true for $n=k$. It is true for $n=1$, so by mathematical induction it is true for $n=1,2,3,\ldots$

The marker needs supporting arguments set out as a proof, and any equivalently worded conclusion is accepted. An equivalent method of proof earns the same marks, such as factorising terms using $1-(-1)^{k+1}$ or working from the right-hand side to the left-hand side.

Marker: completes proof and communicates a suitable conclusion

Putting it all together

Each phrase of the question gave you something

“Given $\cos(\theta)\neq0$”
The right-hand side divides by $2\cos(\theta)$, so this makes sure it has a value. You don’t need to use it in your working.
“use mathematical induction to prove”
You need the initial statement, the assumption with its target, the inductive step and the conclusion. Checking a few values of $n$ is not a proof.
“$(-1)^{n+1}\cos\big((2n-1)\theta\big)$”
This is the general term. Put $n=k+1$ into it to get the next term, $(-1)^{k+2}\cos\big((2k+1)\theta\big)$.
The right-hand side, $\dfrac{1-(-1)^n\cos(2n\theta)}{2\cos(\theta)}$
The denominator tells you to multiply the next term by $2\cos(\theta)$, which sets up the product identity. The $(-1)^n$ tells you the signs will need tidying with powers of $-1$.
What makes this complex familiar

The question gives you one instruction and six marks, with no parts to guide you through the inductive step. You have to write the next term yourself, then spot that $2\cos(\theta)\cos\big((2k+1)\theta\big)$ needs the product identity, and keep track of three different powers of $-1$. It is not complex unfamiliar, because the question names the method and the four parts of an induction proof are ones you have practised many times.

If I had to guess where you might lose marks, it would be the sign of the next term. Writing $(-1)^{k+1}$ instead of $(-1)^{k+2}$ stops the $\cos(2k\theta)$ terms from cancelling, and the working in steps 4 to 6 can no longer reach the right-hand side. The other risk is mixing the two sides together. For this exam, the subject report lists among its practices to strengthen “completing proofs of results for sums for any positive integer using mathematical induction by ensuring algebraic procedures involving one side of the proof are conducted independently of the other side”. Work on the left-hand side alone until it becomes the right-hand side.

Only mark this done when you could do it without help. Reading the solution does not count.

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Question wording and marking-guide steps are from the 2020 QCAA Specialist Mathematics external assessment, © State of Queensland (QCAA) 2020, licensed under CC BY 4.0, and have been adapted. Tangent Tuition is not affiliated with the QCAA.

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