Year 11 Maths Advanced Module 2 Module Quiz

Module 2 Quiz — Trigonometric Functions and Graphs

A comprehensive quiz covering all topics from Module 2: exact values, identities, reciprocal functions, domains and ranges, graphing, transformations, solving equations, and modelling.

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MC

Multiple Choice

10 random questions from a replayable module quiz bank

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Extended Questions

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11. If $\sin \theta = -\frac{2}{3}$ and $\pi < \theta < \frac{3\pi}{2}$, find the exact values of $\cos \theta$ and $\tan \theta$. 3 MARKS

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12. Prove that $1 + \tan^2 \theta = \sec^2 \theta$ starting from $\sin^2 \theta + \cos^2 \theta = 1$. State any restrictions. 3 MARKS

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13. For $y = 3\cos\left(2x + \frac{\pi}{2}\right) - 1$, state the amplitude, period, phase shift, and range. 4 MARKS

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14. Sketch $y = \tan x$ and $y = 1$ on the same axes for $-\frac{\pi}{2} < x < \frac{3\pi}{2}$. Hence, solve $\tan x = 1$ in this interval. 3 MARKS

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15. A Ferris wheel has a radius of 15 metres and its centre is 18 metres above the ground. It completes one full rotation every 6 minutes. A passenger gets on at the lowest point. Let $h$ be the height of the passenger $t$ minutes after getting on. (a) Write a model for $h$ in terms of $t$. (b) Find the height after 1.5 minutes. (c) Find the first time when the passenger is 30 metres above the ground. 6 MARKS

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16. A student writes: "The equation $\sin x = 0.8$ has exactly two solutions because the sine graph is a wave that crosses any horizontal line twice." Evaluate this statement, correcting any errors and explaining when the statement would be true or false. 3 MARKS

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✅ Comprehensive Answers

❓ Multiple Choice

1. A — $\sin \frac{3\pi}{4} = \frac{\sqrt{2}}{2}$.

2. A — $\csc x = \frac{1}{\sin x}$.

3. A — $1 + \tan^2 x = \sec^2 x$.

4. A — $\sin 70^\circ = \cos 20^\circ$.

5. A — Tangent domain excludes odd multiples of $\frac{\pi}{2}$.

6. A — Period of $\cos(3x)$ is $\frac{2\pi}{3}$.

7. A — Range of $2\sin x + 3$ is $[1, 5]$.

8. A — Phase shift = $\frac{\pi}{4}$ right.

9. A — $\tan x$ has vertical asymptotes at $x = \frac{\pi}{2} + n\pi$.

10. A — $a = \frac{12 - 4}{2} = 4$.

📝 Short Answer Model Answers

Q11 (3 marks): $\cos^2 \theta = 1 - \frac{4}{9} = \frac{5}{9}$ [1]. In QIII, $\cos \theta = -\frac{\sqrt{5}}{3}$ [1]. $\tan \theta = \frac{-\frac{2}{3}}{-\frac{\sqrt{5}}{3}} = \frac{2}{\sqrt{5}} = \frac{2\sqrt{5}}{5}$ [1].

Q12 (3 marks): Start with $\sin^2 \theta + \cos^2 \theta = 1$ [0.5]. Divide by $\cos^2 \theta$ (provided $\cos \theta \neq 0$) [0.5]. This gives $\tan^2 \theta + 1 = \sec^2 \theta$ [2].

Q13 (4 marks): Rewrite: $y = 3\cos\left(2\left(x + \frac{\pi}{4}\right)\right) - 1$ [1]. Amplitude = 3 [0.5], Period = $\pi$ [1], Phase shift = $\frac{\pi}{4}$ left [1], Range = $[-4, 2]$ [0.5].

Q14 (3 marks): Sketch of $y = \tan x$ with asymptotes at $x = -\frac{\pi}{2}, \frac{\pi}{2}, \frac{3\pi}{2}$ and horizontal line $y = 1$ [1]. Each branch intersects $y = 1$ once [1]. Solutions: $x = \frac{\pi}{4}, \frac{5\pi}{4}$ [1].

Q15 (6 marks): (a) $a = 15$, $d = 18$, $b = \frac{\pi}{3}$. Starts at minimum, so $h = 18 - 15\cos\left(\frac{\pi}{3}t\right)$ or $h = 18 + 15\sin\left(\frac{\pi}{3}(t - 1.5)\right)$ [2]. (b) At $t = 1.5$: $h = 18 - 15\cos\left(\frac{\pi}{2}\right) = 18$ m [2]. (c) $18 - 15\cos\left(\frac{\pi}{3}t\right) = 30 \Rightarrow \cos\left(\frac{\pi}{3}t\right) = -0.8 \Rightarrow \frac{\pi}{3}t = \cos^{-1}(-0.8) \approx 2.214 \Rightarrow t \approx 2.12$ min [2].

Q16 (3 marks): The statement is incorrect because it ignores the domain [1]. In one period ($2\pi$), $\sin x = 0.8$ has two solutions, but over all real numbers there are infinitely many [1]. The statement is only true when the domain is exactly one period long and $0.8$ is within the range of sine [1].