Comprehensive assessment covering reciprocal trigonometric functions, the Pythagorean and quotient identities, exact values, graphs of sine, cosine and tangent, amplitude and period, and solving trigonometric equations over a given domain.
Assessment
Select the best answer for each question. 1 mark each.
The reciprocal function $\sec\theta$ is defined as:
Simplify $\sin^2\theta + \cos^2\theta$:
The identity $1 + \tan^2\theta$ is equal to:
The exact value of $\sin 30°$ is:
The quotient identity states that $\tan\theta$ equals:
The period of $y = \sin x$ is:
The amplitude of $y = 3\sin(2x)$ is:
The period of $y = \cos(3x)$ is:
The reciprocal function $\csc\theta$ (that is, $\operatorname{cosec}\theta$) is defined as:
Solving $\cos x = \frac{1}{2}$ for $0 \leq x \leq 2\pi$ gives:
Simplify $1 - \cos^2\theta$:
The exact value of $\tan 60°$ is:
The maximum value of $y = \sin x$ is:
Solving $\tan x = 1$ for $0 \leq x \leq 2\pi$ gives:
Which expression is equivalent to $\cot\theta$?
Short Answer
Prove the identity $\dfrac{\sin\theta}{1 + \cos\theta} + \dfrac{1 + \cos\theta}{\sin\theta} = \dfrac{2}{\sin\theta}$.
Simplify $\dfrac{1 - \sin^2\theta}{\cos\theta}$, showing your working.
Solve $2\sin x - 1 = 0$ for $0 \leq x \leq 2\pi$, giving exact answers.
Sketch $y = 2\cos x$ for $0 \leq x \leq 2\pi$, stating the amplitude and period and marking all intercepts.
Solve $\tan^2 x = 3$ for $0 \leq x \leq \pi$, giving exact answers.
Q1: B, $\sec\theta = \frac{1}{\cos\theta}$ by definition.
Q2: D, the Pythagorean identity gives $\sin^2\theta + \cos^2\theta = 1$.
Q3: A, dividing $\sin^2\theta + \cos^2\theta = 1$ by $\cos^2\theta$ gives $\tan^2\theta + 1 = \sec^2\theta$.
Q4: C, $\sin 30° = \frac{1}{2}$.
Q5: A, the quotient identity is $\tan\theta = \frac{\sin\theta}{\cos\theta}$.
Q6: D, $y = \sin x$ repeats every $2\pi$.
Q7: C, amplitude is the coefficient of the trig term, $|3| = 3$.
Q8: B, period $= \frac{2\pi}{3}$ since period $= \frac{2\pi}{b}$ with $b = 3$.
Q9: D, $\csc\theta = \frac{1}{\sin\theta}$ by definition.
Q10: A, $\cos x = \frac{1}{2}$ at $x = \frac{\pi}{3}$ (Q1) and $x = 2\pi - \frac{\pi}{3} = \frac{5\pi}{3}$ (Q4).
Q11: B, rearranging the Pythagorean identity: $1 - \cos^2\theta = \sin^2\theta$.
Q12: D, $\tan 60° = \sqrt{3}$.
Q13: A, the sine curve oscillates between $-1$ and $1$, so the maximum is $1$.
Q14: C, $\tan x = 1$ at $x = \frac{\pi}{4}$ and $x = \frac{\pi}{4} + \pi = \frac{5\pi}{4}$.
Q15: B, $\cot\theta = \frac{1}{\tan\theta} = \frac{\cos\theta}{\sin\theta}$.
Q16 (3 marks): Common denominator: $\frac{\sin^2\theta + (1 + \cos\theta)^2}{\sin\theta(1 + \cos\theta)}$ [1]. Expand: $\sin^2\theta + 1 + 2\cos\theta + \cos^2\theta = (\sin^2\theta + \cos^2\theta) + 1 + 2\cos\theta = 2 + 2\cos\theta$ [1]. So the expression $= \frac{2(1 + \cos\theta)}{\sin\theta(1 + \cos\theta)} = \frac{2}{\sin\theta}$ [1].
Q17 (2 marks): $1 - \sin^2\theta = \cos^2\theta$ [1]. So $\frac{\cos^2\theta}{\cos\theta} = \cos\theta$ [1].
Q18 (3 marks): $\sin x = \frac{1}{2}$ [1]. Reference angle $\frac{\pi}{6}$; sine is positive in quadrants 1 and 2 [1]. So $x = \frac{\pi}{6}, \frac{5\pi}{6}$ [1].
Q19 (3 marks): Amplitude $= 2$, period $= 2\pi$ [1]. Key values: $y(0) = 2$, $y(\frac{\pi}{2}) = 0$, $y(\pi) = -2$, $y(\frac{3\pi}{2}) = 0$, $y(2\pi) = 2$ [1]. $x$-intercepts at $x = \frac{\pi}{2}$ and $x = \frac{3\pi}{2}$; smooth cosine curve through these points [1].
Q20 (3 marks): $\tan x = \pm\sqrt{3}$ [1]. On $[0, \pi]$: $\tan x = \sqrt{3}$ gives $x = \frac{\pi}{3}$; $\tan x = -\sqrt{3}$ gives $x = \frac{2\pi}{3}$ [1]. So $x = \frac{\pi}{3}, \frac{2\pi}{3}$ [1].