QCAA Specialist Mathematics Alternative Sequence Trigonometry and functions

4 sample questions with marking guides and sample answers

Q18
2022
QCAA
Paper 1
6 marks
Q18
6 marks

Points in a plane are transformed by a rotation about the origin through angle AA followed by a rotation in the opposite direction about the origin through angle BB.

Use the composition of these transformations to prove
sin⁡(A−B)=sin⁡(A)cos⁡(B)−cos⁡(A)sin⁡(B)\sin(A - B) = \sin(A)\cos(B) - \cos(A)\sin(B)
cos⁡(A−B)=cos⁡(A)cos⁡(B)+sin⁡(A)sin⁡(B)\cos(A - B) = \cos(A)\cos(B) + \sin(A)\sin(B)

Reveal Answer

Method 1: Assumes the rotation through A as the positive direction.
First rotation matrix:
RA=[cos⁡(A)−sin⁡(A)sin⁡(A)cos⁡(A)]\mathbf{R}_A = \begin{bmatrix} \cos(A) & -\sin(A) \\ \sin(A) & \cos(A) \end{bmatrix}

Second rotation matrix:
R−B=[cos⁡(−B)−sin⁡(−B)sin⁡(−B)cos⁡(−B)]\mathbf{R}_{-B} = \begin{bmatrix} \cos(-B) & -\sin(-B) \\ \sin(-B) & \cos(-B) \end{bmatrix}
=[cos⁡(B)sin⁡(B)−sin⁡(B)cos⁡(B)]= \begin{bmatrix} \cos(B) & \sin(B) \\ -\sin(B) & \cos(B) \end{bmatrix}

Matrix representing the composition of the successive rotations:
R−BRA=[cos⁡(B)sin⁡(B)−sin⁡(B)cos⁡(B)][cos⁡(A)−sin⁡(A)sin⁡(A)cos⁡(A)]\mathbf{R}_{-B} \mathbf{R}_A = \begin{bmatrix} \cos(B) & \sin(B) \\ -\sin(B) & \cos(B) \end{bmatrix} \begin{bmatrix} \cos(A) & -\sin(A) \\ \sin(A) & \cos(A) \end{bmatrix}
=[cos⁡(B)cos⁡(A)+sin⁡(B)sin⁡(A)−cos⁡(B)sin⁡(A)+sin⁡(B)cos⁡(A)−sin⁡(B)cos⁡(A)+cos⁡(B)sin⁡(A)sin⁡(B)sin⁡(A)+cos⁡(B)cos⁡(A)]= \begin{bmatrix} \cos(B)\cos(A) + \sin(B)\sin(A) & -\cos(B)\sin(A) + \sin(B)\cos(A) \\ -\sin(B)\cos(A) + \cos(B)\sin(A) & \sin(B)\sin(A) + \cos(B)\cos(A) \end{bmatrix}

Matrix representing transformation of the combined rotation of A−BA - B
RA−B=[cos⁡(A−B)−sin⁡(A−B)sin⁡(A−B)cos⁡(A−B)]\mathbf{R}_{A-B} = \begin{bmatrix} \cos(A - B) & -\sin(A - B) \\ \sin(A - B) & \cos(A - B) \end{bmatrix}

R−BRA\mathbf{R}_{-B} \mathbf{R}_A and RA−B\mathbf{R}_{A-B} represent the same transformation.

Equating parts:
cos⁡(A−B)=cos⁡(A)cos⁡(B)+sin⁡(A)sin⁡(B)\cos(A - B) = \cos(A)\cos(B) + \sin(A)\sin(B)
sin⁡(A−B)=sin⁡(A)cos⁡(B)−cos⁡(A)sin⁡(B)\sin(A - B) = \sin(A)\cos(B) - \cos(A)\sin(B)

Marking Criteria
DescriptorMarks

correctly determines the matrix representing the first rotation in terms of angle AA

1

correctly determines the matrix representing the second rotation in terms of angle BB

1

determines composition of successive rotations using a matrix product

1

determines a matrix that represents the composition of the two transformations in terms of angles AA and BB

1

correctly determines a matrix that represents the composition of the two transformations in terms of angle (A−B)(A - B)

1

uses mathematical reasoning to complete the proof

1
Q16
2021
QCAA
Paper 2
6 marks
Q16

The expression (32−3)sin⁡(x)−3cos⁡(x)(3\sqrt{2} - 3)\sin(x) - \sqrt{3}\cos(x) can be converted into the form Rcos⁡(x+α)R\cos(x + \alpha), where R>0R > 0 and 0<α<2π0 < \alpha < 2\pi.

Q16a
4 marks

Determine the values of RR and α\alpha.

Reveal Answer

(32−3)sin⁡(x)−3cos⁡(x)=Rcos⁡(x+α)(3\sqrt{2} - 3) \sin(x) - \sqrt{3} \cos(x) = R \cos(x + \alpha)

=Rcos⁡(x)cos⁡(α)−Rsin⁡(x)sin⁡(α)= R \cos(x) \cos(\alpha) - R \sin(x) \sin(\alpha)

Equating parts
Rsin⁡(α)=−(32−3)...(1)R \sin(\alpha) = -(3\sqrt{2} - 3) \quad ... (1)
Rcos⁡(α)=−3...(2)R \cos(\alpha) = -\sqrt{3} \quad ... (2)

Solving simultaneously
Rsin⁡(α)Rcos⁡(α)=32−33\frac{R \sin(\alpha)}{R \cos(\alpha)} = \frac{3\sqrt{2} - 3}{\sqrt{3}}

α=tan⁡−1(32−33)\alpha = \tan^{-1} \left( \frac{3\sqrt{2} - 3}{\sqrt{3}} \right)

α≈π+0.62≈3.76\alpha \approx \pi + 0.62 \approx 3.76
(as sin⁡(α)<0,cos⁡(α)<0\sin(\alpha) < 0, \cos(\alpha) < 0)

Substituting into (2)
R≈−−3cos⁡(3.764)≈2.13R \approx -\frac{-\sqrt{3}}{\cos(3.764)} \approx 2.13

Marking Criteria
DescriptorMarks

correctly forms two simultaneous equations in terms of RR and α\alpha

1

uses a suitable technique to solve the simultaneous equations

1

solves for α\alpha by considering the suitable quadrant

1

solves for RR

1
Q16b
2 marks

Use the results from Question 16a) or another method to solve the equation
Rcos⁡(x+α)=0.5R\cos(x + \alpha) = 0.5 where 0≤x≤2π0 \leq x \leq 2\pi.

Reveal Answer

Solving 2.13cos⁡(x+α)=0.52.13 \cos(x + \alpha) = 0.5 using GDC
x≈1.19x \approx 1.19

x≈3.85x \approx 3.85

Marking Criteria
DescriptorMarks

determines solution in a suitable quadrant of the given domain

1

determines solution in another suitable quadrant of the given domain

1
Q7
2021
QCAA
Paper 1
1 mark
Q7
1 mark

The product 2cos⁡(25∘)cos⁡(75∘)2\cos(25^\circ)\cos(75^\circ) can be expressed as

A

cos⁡(50∘)+cos⁡(80∘)\cos(50^\circ)+\cos(80^\circ)

B

cos⁡(50∘)−cos⁡(80∘)\cos(50^\circ)-\cos(80^\circ)

C

−cos⁡(50∘)+cos⁡(80∘)-\cos(50^\circ)+\cos(80^\circ)

D

−cos⁡(50∘)−cos⁡(80∘)-\cos(50^\circ)-\cos(80^\circ)

Reveal Answer
A

cos⁡(50∘)+cos⁡(80∘)\cos(50^\circ)+\cos(80^\circ)

The product-to-sum formula yields cos⁡(100∘)+cos⁡(50∘)\cos(100^\circ) + \cos(50^\circ). Since cos⁡(100∘)=−cos⁡(80∘)\cos(100^\circ) = -\cos(80^\circ), the sign for the 80∘80^\circ term must be negative, not positive.

B

cos⁡(50∘)−cos⁡(80∘)\cos(50^\circ)-\cos(80^\circ)

Correct Answer

Applying the product-to-sum formula 2cos⁡(A)cos⁡(B)=cos⁡(A−B)+cos⁡(A+B)2\cos(A)\cos(B) = \cos(A-B) + \cos(A+B) gives cos⁡(−50∘)+cos⁡(100∘)\cos(-50^\circ) + \cos(100^\circ). Since cos⁡(−50∘)=cos⁡(50∘)\cos(-50^\circ) = \cos(50^\circ) and cos⁡(100∘)=−cos⁡(80∘)\cos(100^\circ) = -\cos(80^\circ), this simplifies to cos⁡(50∘)−cos⁡(80∘)\cos(50^\circ) - \cos(80^\circ).

C

−cos⁡(50∘)+cos⁡(80∘)-\cos(50^\circ)+\cos(80^\circ)

This expression has the opposite signs of the correct result, which would correspond to the product −2cos⁡(25∘)cos⁡(75∘)-2\cos(25^\circ)\cos(75^\circ).

D

−cos⁡(50∘)−cos⁡(80∘)-\cos(50^\circ)-\cos(80^\circ)

This expression evaluates to −cos⁡(50∘)+cos⁡(100∘)-\cos(50^\circ) + \cos(100^\circ), which does not match the result of the product-to-sum formula for the given expression.

Q16
2020
QCAA
Paper 2
6 marks
Q16

Consider the identity

cos⁡(4θ)=Acos⁡4(θ)+Bsin⁡2(θ)+C\cos(4\theta) = A\cos^4(\theta) + B\sin^2(\theta) + C where A,BA, B and C∈ZC \in \mathbb{Z}

Q16a
5 marks

Determine the values of A,BA, B and CC using De Moivre's theorem.

Reveal Answer

Using De Moivre’s theorem:\text{Using De Moivre's theorem:} (cis(θ))4=cis(4θ)(\text{cis}(\theta))^4 = \text{cis}(4\theta) Equating real parts\text{Equating real parts} cos⁡(4θ)=Re(cos⁡(θ)+isin⁡(θ))4\cos(4\theta) = Re(\cos(\theta) + i\sin(\theta))^4 =cos⁡4(θ)−6cos⁡2(θ)sin⁡2(θ)+sin⁡4(θ)= \cos^4(\theta) - 6\cos^2(\theta)\sin^2(\theta) + \sin^4(\theta) =cos⁡4(θ)−6cos⁡2(θ)(1−cos⁡2(θ))+(1−cos⁡2(θ))2= \cos^4(\theta) - 6\cos^2(\theta)(1 - \cos^2(\theta)) + (1 - \cos^2(\theta))^2 =cos⁡4(θ)−6cos⁡2(θ)+6cos⁡4(θ)+1−2cos⁡2(θ)+cos⁡4(θ)= \cos^4(\theta) - 6\cos^2(\theta) + 6\cos^4(\theta) + 1 - 2\cos^2(\theta) + \cos^4(\theta) =8cos⁡4(θ)−8cos⁡2(θ)+1= 8\cos^4(\theta) - 8\cos^2(\theta) + 1 =8cos⁡4(θ)−8(1−sin⁡2(θ))+1= 8\cos^4(\theta) - 8(1 - \sin^2(\theta)) + 1 =8cos⁡4(θ)−8+8sin⁡2(θ)+1= 8\cos^4(\theta) - 8 + 8\sin^2(\theta) + 1 =8cos⁡4(θ)+8sin⁡2(θ)−7= 8\cos^4(\theta) + 8\sin^2(\theta) - 7 So A=8,B=8,C=−7\text{So } A = 8, B = 8, C = -7
Marking Criteria
DescriptorMarks

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 Acos⁡4(θ)+Bsin⁡2(θ)+CA\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

A verification strategy would be to graph y=cos⁡(4θ) and y=8cos⁡4(θ)+8sin⁡2(θ)−7 to confirm that the two graphs are the same.\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
DescriptorMarks

describes an appropriate verification strategy

1

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