Test your understanding of reciprocal trig functions, Pythagorean identities, complementary angles, and domains and ranges.
8 random questions from a replayable checkpoint bank
✍️ Short Answer
9. If $\tan \theta = \frac{3}{4}$ and $\pi < \theta < \frac{3\pi}{2}$, find the exact values of $\sin \theta$, $\cos \theta$, and $\sec \theta$. 3 MARKS
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10. Prove that $1 + \cot^2 \theta = \csc^2 \theta$ starting from $\sin^2 \theta + \cos^2 \theta = 1$. State any restrictions. 3 MARKS
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11. Solve $\sin 2\theta = \cos \theta$ for $0^\circ \leq \theta \leq 90^\circ$, giving exact answers where possible. 3 MARKS
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12. (a) State the domain and range of $y = \csc x$. (b) Find all values of $x$ in $[0, 2\pi]$ where $\csc x$ is undefined. (c) Explain why the range of $y = \csc x$ does not include values between $-1$ and $1$. 4 MARKS
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1. A — $\csc x = \frac{1}{\sin x}$.
2. A — $\sec^2 x - \tan^2 x = 1$.
3. A — $\sin 40^\circ = \cos 50^\circ$.
4. A — Tangent undefined at odd multiples of $\frac{\pi}{2}$.
5. A — Secant range: $y \leq -1$ or $y \geq 1$.
6. A — $\cot x = \frac{1}{\tan x} = \frac{3}{4}$.
7. A — $\cos^2 x = 1 - \sin^2 x$.
8. A — $2\sin x - 1$ has range $[-3, 1]$.
Q9 (3 marks): Using $1 + \tan^2 \theta = \sec^2 \theta$: $\sec^2 \theta = 1 + \frac{9}{16} = \frac{25}{16}$ [0.5]. In QIII, $\sec \theta < 0$, so $\sec \theta = -\frac{5}{4}$ [0.5], $\cos \theta = -\frac{4}{5}$ [1]. $\sin \theta = \tan \theta \cdot \cos \theta = \frac{3}{4} \times (-\frac{4}{5}) = -\frac{3}{5}$ [1].
Q10 (3 marks): Start with $\sin^2 \theta + \cos^2 \theta = 1$ [0.5]. Divide by $\sin^2 \theta$ (provided $\sin \theta \neq 0$) [0.5]: $1 + \cot^2 \theta = \csc^2 \theta$ [2].
Q11 (3 marks): $\cos \theta = \sin(90^\circ - \theta)$ [1]. So $\sin 2\theta = \sin(90^\circ - \theta)$ [0.5], giving $2\theta = 90^\circ - \theta$ [0.5]. Hence $\theta = 30^\circ$ [1].
Q12 (4 marks): (a) Domain: $x \neq n\pi$, $n \in \mathbb{Z}$. Range: $y \leq -1$ or $y \geq 1$ [1]. (b) $x = 0, \pi, 2\pi$ [1]. (c) $\csc x = \frac{1}{\sin x}$ [0.5]. Since $|\sin x| \leq 1$ and $\sin x \neq 0$, the reciprocal satisfies $|\csc x| \geq 1$ [1.5].