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QCAA Specialist Mathematics Alternative Sequence — Mathematical induction and trigonometric proofs Questions & Answers

Q18

2021

QCAA

Paper 1

6 marks

Q18

6 marks

For $n\in\mathbb{Z}^{+}$ and suitable values of $\theta\in\mathbb{R}$ , prove that

$\sum_{r=1}^{n}2^{r-1}\tan\left(2^{r-1}\theta\right)=\cot(\theta)-2^{n}\cot\left(2^{n}\theta\right)$

Reveal Answer

RTP
$\sum_{r=1}^n 2^{r-1}\tan(2^{r-1}\theta)$
$= \cot(\theta) - 2^n\cot(2^n\theta) \;\forall\; n \in Z^+$
Mathematical induction can be used to prove this proposition.

Initial statement:
Let $n = 1$ :
RTP
$\tan(\theta) = \cot(\theta) - 2\cot(2\theta)$
RHS $= \cot(\theta) - 2\cot(2\theta)$
$= \frac{\cos(\theta)}{\sin(\theta)} - \frac{2\cos(2\theta)}{\sin(2\theta)}$
$= \frac{\cos(\theta)}{\sin(\theta)} - \frac{2(\cos^2(\theta) - \sin^2(\theta))}{2\sin(\theta)\cos(\theta)}$
$= \frac{\cos^2(\theta) - \cos^2(\theta) + \sin^2(\theta)}{\sin(\theta)\cos(\theta)}$
$= \frac{\sin^2(\theta)}{\sin(\theta)\cos(\theta)}$
$= \tan(\theta)$
$=$ LHS
Proposition is true for $n = 1$

Assumption for inductive step
Assume proposition is true for $n = k \;\forall\; k \in Z^+$
$\sum_{r=1}^k 2^{r-1}\tan(2^{r-1}\theta) = \cot(\theta) - 2^k\cot(2^k\theta)$

Inductive step:
Let $n = k + 1$
RTP
$\sum_{r=1}^{k+1} 2^{r-1}\tan(2^{r-1}\theta) = \cot(\theta) - 2^{k+1}\cot(2^{k+1}\theta)$
LHS $= \sum_{r=1}^{k+1} 2^{r-1}\tan(2^{r-1}\theta)$
$= \sum_{r=1}^k 2^{r-1}\tan(2^{r-1}\theta) + 2^k\tan(2^k\theta)$
$= \cot(\theta) - 2^k\cot(2^k\theta) + 2^k\tan(2^k\theta)$
$= \cot(\theta) - 2^k(\cot(2^k\theta) - 2^k\tan(2^k\theta))$
$= \cot(\theta) - 2^k\left(\frac{\cos(2^k\theta)}{\sin(2^k\theta)} - \frac{\sin(2^k\theta)}{\cos(2^k\theta)}\right)$
$= \cot(\theta) - 2^k\left(\frac{\cos^2(2^k\theta) - \sin^2(2^k\theta)}{\sin(2^k\theta)\cos(2^k\theta)}\right)$
$= \cot(\theta) - 2^k\left(\frac{\cos(2(2^k\theta))}{\frac{1}{2}\sin(2(2^k\theta))}\right)$
$= \cot(\theta) - 2 \times 2^k\left(\frac{\cos(2(2^k\theta))}{\sin(2(2^k\theta))}\right)$
$= \cot(\theta) - 2^{k+1}(\cot(2^{k+1}\theta))$
$=$ RHS
So $P(k + 1)$ is true. By mathematical induction, the proposition is true for $n = 1, 2, \dots$

Marking Criteria

Descriptor	Marks
correctly expresses the RHS in terms of $\theta$	1
correctly states a proposition for $n = k$	1
correctly states an inductive step proof requirement	1
substitutes assumption within the inductive step	1
establishes expression providing an opportunity to apply two double-angle trigonometric identities	1
shows evidence of application of a double-angle trigonometric identity to complete the proof	1

Q16

2020

QCAA

Paper 2

6 marks

Q16

Consider the identity

$\cos(4\theta) = A\cos^4(\theta) + B\sin^2(\theta) + C$ where $A, B$ and $C \in \mathbb{Z}$

Q16a

5 marks

Determine the values of $A, B$ and $C$ using De Moivre's theorem.

Reveal Answer

\text{Using De Moivre's theorem:}

(\text{cis}(\theta))^4 = \text{cis}(4\theta)

\text{Equating real parts}

\cos(4\theta) = Re(\cos(\theta) + i\sin(\theta))^4

= \cos^4(\theta) - 6\cos^2(\theta)\sin^2(\theta) + \sin^4(\theta)

= \cos^4(\theta) - 6\cos^2(\theta)(1 - \cos^2(\theta)) + (1 - \cos^2(\theta))^2

= \cos^4(\theta) - 6\cos^2(\theta) + 6\cos^4(\theta) + 1 - 2\cos^2(\theta) + \cos^4(\theta)

= 8\cos^4(\theta) - 8\cos^2(\theta) + 1

= 8\cos^4(\theta) - 8(1 - \sin^2(\theta)) + 1

= 8\cos^4(\theta) - 8 + 8\sin^2(\theta) + 1

= 8\cos^4(\theta) + 8\sin^2(\theta) - 7

\text{So } A = 8, B = 8, C = -7

Marking Criteria

Descriptor	Marks
correctly uses De Moivre's theorem	1
uses binomial expansion with the real parts and simplifies the expression	1
establishes a simplified expression following the use of a suitable Pythagorean identity	1
establishes a simplified expression in the form of $A\cos^4(\theta) + B\sin^2(\theta) + C$	1
communicates the values of A, B and C	1

Q16b

1 mark

State an appropriate method of verifying your results from 16a).

Reveal Answer

\text{A verification strategy would be to graph } y = \cos(4\theta) \text{ and } y = 8\cos^4(\theta) + 8\sin^2(\theta) - 7 \text{ to confirm that the two graphs are the same.}

Marking Criteria

Descriptor	Marks
describes an appropriate verification strategy	1

Q3

2021

QCAA

Paper 2

1 mark

Q3

1 mark

Given $n \in Z^+$ , for which proposition can the initial statement for mathematical induction be proven?

A

$x^{2n} - y^{2n}$ is divisible by $(x+y) \forall (x+y) \neq 0$

B

$1^2 + 2^2 + 3^2 + \dots + n^2 = \frac{1}{6}n(2n^2 + 3n - 1)$

C

$(n+1)^3 + (n+2)^3$ is divisible by 3

D

$\sum_{r=1}^n \frac{1}{(2r-1)(2r+1)} = \frac{n}{n+1}$

Reveal Answer

A

$x^{2n} - y^{2n}$ is divisible by $(x+y) \forall (x+y) \neq 0$

Correct Answer

For the base case $n=1$ , the expression becomes $x^2 - y^2 = (x-y)(x+y)$ , which is clearly divisible by $(x+y)$ . Thus, the initial statement is true.

B

$1^2 + 2^2 + 3^2 + \dots + n^2 = \frac{1}{6}n(2n^2 + 3n - 1)$

For $n=1$ , the left side is $1^2 = 1$ , but the right side evaluates to $\frac{1}{6}(1)(2(1)^2 + 3(1) - 1) = \frac{2}{3}$ . Since $1 \neq \frac{2}{3}$ , the base case fails.

C

$(n+1)^3 + (n+2)^3$ is divisible by 3

For $n=1$ , the expression evaluates to $(1+1)^3 + (1+2)^3 = 2^3 + 3^3 = 8 + 27 = 35$ . Since 35 is not divisible by 3, the base case fails.

D

$\sum_{r=1}^n \frac{1}{(2r-1)(2r+1)} = \frac{n}{n+1}$

For $n=1$ , the left side evaluates to $\frac{1}{(1)(3)} = \frac{1}{3}$ , while the right side evaluates to $\frac{1}{1+1} = \frac{1}{2}$ . Since $\frac{1}{3} \neq \frac{1}{2}$ , the base case fails.

Q13

2021

QCAA

Paper 1

7 marks

Q13

The proposition that $2^{2n}+3n-1$ is divisible by 3 $\forall\, n\in\mathbb{Z}^{+}$ can be restated as

$P(n): 2^{2n}+3n-1=3m\ \forall\, n\in\mathbb{Z}^{+}$ , for some $m\in\mathbb{Z}^{+}$ .

Complete the proof by mathematical induction steps to prove that $P(n)$ is true.

Q13a

1 mark

Prove that $P(1)$ is true.

Reveal Answer

RTP
$P(n): 2^{2n} + 3n - 1 = 3m \;\forall\; n \in Z^+$ , for some $m \in Z^+$
$P(1): 2^2 + 3 - 1 = 6$
$= 3 \times 2$
Proposition is true for $n = 1$

Marking Criteria

Descriptor	Marks
correctly proves the initial statement	1

Q13b

1 mark

State $P(k)$ .

Reveal Answer

$P(k): 2^{2k} + 3k - 1 = 3m \;\forall\; n \in Z^+$ for some $m \in Z^+$

Marking Criteria

Descriptor	Marks
correctly states $P(k)$	1

Q13c

4 marks

Use the assumption that $P(k)$ is true $\forall\, k\in\mathbb{Z}^{+}$ to prove that $P(k+1)$ is true.

Reveal Answer

Prove $P(k + 1)$ is true assuming $P(k)$ is true $\forall\; k \in Z^+$ .
$2^{2(k+1)} + 3(k + 1) - 1$
$= 2^{2k+2} + 3k + 3 - 1$
$= 2^2 \times 2^{2k} + 3k - 1 + 3$
$= 2^{2k} + 3k - 1 + 3 \times 2^{2k} + 3$
$= 3m + 3 \times 2^{2k} + 3$
$= 3(m + 2^{2k} + 1)$
$= 3p$ for some $p \in Z^+$

Marking Criteria

Descriptor	Marks
correctly establishes an expression representing the LHS of $P(k + 1)$	1
uses an index law to establish a term with a factor of $2^{2k}$	1
uses assumption in the inductive step	1
completes proof by determining an expression with a factor of 3 representing the RHS of $P(k + 1)$	1

Q13d

1 mark

State a conclusion to the proof.

Reveal Answer

Conclusion
So $P(k + 1)$ is true.
By mathematical induction, the proposition is true for $n = 1, 2, \dots$

Marking Criteria

Descriptor	Marks
communicates a suitable conclusion having attempted to complete Question 13c)	1

QCAA Specialist Mathematics Alternative Sequence Mathematical induction and trigonometric proofs

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