A Level H2 Mathematics Geometry Trigonometry Quiz

A-Level Maths H2 Quiz - Geometry Trigonometry: ANSWER KEY

Total Marks: 60

Section A: Trigonometric Functions and Equations (20 marks)

1. Given $\sin \theta = \frac{3}{5}$, $\theta$ acute. $\cos \theta = \sqrt{1 - \sin^2 \theta} = \sqrt{1 - \frac{9}{25}} = \sqrt{\frac{16}{25}} = \frac{4}{5}$ [M1] $\sec \theta = \frac{1}{\cos \theta} = \frac{5}{4}$ [A1] [2 marks]

2. $2\cos^2 x - 3\sin x - 3 = 0$, $0^\circ \le x \le 360^\circ$. Using $\cos^2 x = 1 - \sin^2 x$: $2(1 - \sin^2 x) - 3\sin x - 3 = 0$ $2 - 2\sin^2 x - 3\sin x - 3 = 0$ $-2\sin^2 x - 3\sin x - 1 = 0$ $2\sin^2 x + 3\sin x + 1 = 0$ [M1] $(2\sin x + 1)(\sin x + 1) = 0$ [M1] $\sin x = -\frac{1}{2}$ or $\sin x = -1$ For $\sin x = -\frac{1}{2}$: $x = 210^\circ, 330^\circ$ For $\sin x = -1$: $x = 270^\circ$ [A1 for all three] $\therefore x = 210^\circ, 270^\circ, 330^\circ$ [A1] [4 marks]

3. Prove $\displaystyle \frac{\sin 2A}{1 + \cos 2A} = \tan A$. LHS $= \frac{2\sin A\cos A}{1 + (2\cos^2 A - 1)}$ [M1: use double angle formulas] $= \frac{2\sin A\cos A}{2\cos^2 A}$ [M1: simplify denominator] $= \frac{\sin A}{\cos A} = \tan A =$ RHS [A1] [3 marks]

4. $y = 3\sin 2x + 4\cos 2x$, $0 \le x \le \pi$.

(a) $R = \sqrt{3^2 + 4^2} = \sqrt{25} = 5$ [M1] $\tan \alpha = \frac{4}{3}$, so $\alpha = \tan^{-1}\left(\frac{4}{3}\right) \approx 0.927$ rad (3 d.p.) [M1] $\therefore y = 5\sin(2x + 0.927)$ [A1]

(b) Maximum value of $y = 5$ [B1] Occurs when $\sin(2x + 0.927) = 1$ $2x + 0.927 = \frac{\pi}{2} + 2k\pi$ Smallest positive $x$: $2x = \frac{\pi}{2} - 0.927 = 1.5708 - 0.9273 = 0.6435$ $x = 0.322$ rad (3 s.f.) [A1] [5 marks]

5. $\tan\left(x + \frac{\pi}{6}\right) = \sqrt{3}$, $0 \le x \le 2\pi$. $\tan\left(x + \frac{\pi}{6}\right) = \sqrt{3} \implies x + \frac{\pi}{6} = \frac{\pi}{3} + k\pi$ [M1] $x = \frac{\pi}{3} - \frac{\pi}{6} + k\pi = \frac{\pi}{6} + k\pi$ [M1] For $k = 0$: $x = \frac{\pi}{6}$ For $k = 1$: $x = \frac{7\pi}{6}$ For $k = 2$: $x = \frac{13\pi}{6} > 2\pi$ (reject) $\therefore x = \frac{\pi}{6}, \frac{7\pi}{6}$ [A1] [3 marks]

6. $\cos \theta = -\frac{5}{13}$, $\pi < \theta < \frac{3\pi}{2}$ (third quadrant). $\sin \theta = -\sqrt{1 - \cos^2 \theta} = -\sqrt{1 - \frac{25}{169}} = -\sqrt{\frac{144}{169}} = -\frac{12}{13}$ [M1] $\sin 2\theta = 2\sin \theta \cos \theta = 2\left(-\frac{12}{13}\right)\left(-\frac{5}{13}\right)$ [M1] $= \frac{120}{169}$ [A1] [3 marks]

Section B: Trigonometric Graphs and Transformations (20 marks)

7. $f(x) = a\cos(bx) + c$, max $(\pi, 5)$, min $(0, 1)$.

(a) Amplitude $= \frac{5 - 1}{2} = 2$, so $a = 2$ [B1] Vertical shift $c = \frac{5 + 1}{2} = 3$ [B1] Period: distance from min to max is half period $= \pi$, so period $= 2\pi$. $b = \frac{2\pi}{\text{period}} = \frac{2\pi}{2\pi} = 1$ [B1] $\therefore a = 2, b = 1, c = 3$

(b) $y = 2f(x) - 1 = 2(2\cos x + 3) - 1 = 4\cos x + 6 - 1 = 4\cos x + 5$ [M1] Max: $4(1) + 5 = 9$ at $x = 0, 2\pi$ [A1] Min: $4(-1) + 5 = 1$ at $x = \pi$ [A1] Sketch: cosine curve with amplitude 4, shifted up 5. [6 marks]

8. $g(x) = \sec x$, $0 \le x \le \pi$, $x \ne \frac{\pi}{2}$.

(a) Asymptote: $x = \frac{\pi}{2}$ [B1]

(b) $y = \sec x = \frac{1}{\cos x}$. $y$-intercept: $x = 0$, $y = \sec 0 = 1$ [B1] As $x \to \frac{\pi}{2}^-$, $\cos x \to 0^+$, so $y \to +\infty$ As $x \to \frac{\pi}{2}^+$, $\cos x \to 0^-$, so $y \to -\infty$ At $x = \pi$: $y = \sec \pi = -1$ [B1] Sketch: U-shaped branch for $0 \le x < \frac{\pi}{2}$ with minimum at $(0, 1)$, approaching $+\infty$ at asymptote; inverted U-shaped branch for $\frac{\pi}{2} < x \le \pi$ with maximum at $(\pi, -1)$, approaching $-\infty$ at asymptote. [A1 for correct sketch] [4 marks]

9. $y = \sin x + \sqrt{3}\cos x$, $0 \le x \le 2\pi$.

(a) $R = \sqrt{1^2 + (\sqrt{3})^2} = \sqrt{4} = 2$ [M1] $\tan \alpha = \frac{\sqrt{3}}{1} = \sqrt{3}$, so $\alpha = \frac{\pi}{3}$ [M1] $\therefore \sin x + \sqrt{3}\cos x = 2\sin\left(x + \frac{\pi}{3}\right)$ [A1]

(b) $C$ meets $x$-axis when $y = 0$: $2\sin\left(x + \frac{\pi}{3}\right) = 0$ $\sin\left(x + \frac{\pi}{3}\right) = 0$ [M1] $x + \frac{\pi}{3} = 0, \pi, 2\pi, 3\pi, \ldots$ $x = -\frac{\pi}{3}, \frac{2\pi}{3}, \frac{5\pi}{3}, \frac{8\pi}{3}, \ldots$ [M1] In domain $0 \le x \le 2\pi$: $x = \frac{2\pi}{3}, \frac{5\pi}{3}$ Coordinates: $\left(\frac{2\pi}{3}, 0\right)$ and $\left(\frac{5\pi}{3}, 0\right)$ [A1] [6 marks]

10. $h(x) = 2\sin\left(3x - \frac{\pi}{4}\right)$, $0 \le x \le \pi$.

(a) Amplitude $= 2$ [B1] Period $= \frac{2\pi}{3}$ [B1]

(b) $h(x) = 1$: $2\sin\left(3x - \frac{\pi}{4}\right) = 1$ $\sin\left(3x - \frac{\pi}{4}\right) = \frac{1}{2}$ [M1] $3x - \frac{\pi}{4} = \frac{\pi}{6} + 2k\pi$ or $3x - \frac{\pi}{4} = \frac{5\pi}{6} + 2k\pi$ $3x = \frac{\pi}{6} + \frac{\pi}{4} + 2k\pi = \frac{5\pi}{12} + 2k\pi$ or $3x = \frac{5\pi}{6} + \frac{\pi}{4} + 2k\pi = \frac{13\pi}{12} + 2k\pi$ $x = \frac{5\pi}{36} + \frac{2k\pi}{3}$ or $x = \frac{13\pi}{36} + \frac{2k\pi}{3}$ For $k = 0$: $x = \frac{5\pi}{36}, \frac{13\pi}{36}$ For $k = 1$: $x = \frac{5\pi}{36} + \frac{2\pi}{3} = \frac{29\pi}{36}, \frac{13\pi}{36} + \frac{2\pi}{3} = \frac{37\pi}{36} > \pi$ (reject) $\therefore x = \frac{5\pi}{36}, \frac{13\pi}{36}, \frac{29\pi}{36}$ [A1] [4 marks]

Section C: Trigonometric Applications and Proofs (20 marks)

11. Triangle $ABC$: $AB = 8$, $AC = 6$, $\angle BAC = 60^\circ$.

(a) Cosine rule: $BC^2 = AB^2 + AC^2 - 2(AB)(AC)\cos 60^\circ$ $= 64 + 36 - 2(8)(6)\left(\frac{1}{2}\right)$ [M1] $= 100 - 48 = 52$ $BC = \sqrt{52} = 2\sqrt{13}$ cm [A1]

(b) Area $= \frac{1}{2}(AB)(AC)\sin 60^\circ$ $= \frac{1}{2}(8)(6)\left(\frac{\sqrt{3}}{2}\right)$ [M1] $= 12\sqrt{3}$ cm$^2$ [A1] [4 marks]

12. $\sin 2\theta = \cos \theta$, $0 \le \theta \le 2\pi$. $2\sin \theta \cos \theta = \cos \theta$ [M1] $2\sin \theta \cos \theta - \cos \theta = 0$ $\cos \theta(2\sin \theta - 1) = 0$ [M1] $\cos \theta = 0$ or $\sin \theta = \frac{1}{2}$ $\cos \theta = 0$: $\theta = \frac{\pi}{2}, \frac{3\pi}{2}$ [A1] $\sin \theta = \frac{1}{2}$: $\theta = \frac{\pi}{6}, \frac{5\pi}{6}$ [A1] $\therefore \theta = \frac{\pi}{6}, \frac{\pi}{2}, \frac{5\pi}{6}, \frac{3\pi}{2}$ [4 marks]

13. Prove $\displaystyle \frac{\cos 3\theta + \cos \theta}{\sin 3\theta - \sin \theta} = \cot \theta$.

Using sum-to-product formulas: $\cos 3\theta + \cos \theta = 2\cos\left(\frac{3\theta + \theta}{2}\right)\cos\left(\frac{3\theta - \theta}{2}\right) = 2\cos 2\theta \cos \theta$ [M1] $\sin 3\theta - \sin \theta = 2\cos\left(\frac{3\theta + \theta}{2}\right)\sin\left(\frac{3\theta - \theta}{2}\right) = 2\cos 2\theta \sin \theta$ [M1]

LHS $= \frac{2\cos 2\theta \cos \theta}{2\cos 2\theta \sin \theta}$ [M1] $= \frac{\cos \theta}{\sin \theta} = \cot \theta =$ RHS [A1] [4 marks]

14. Triangle $PQR$: $PQ = 10$, $PR = 7$, $\angle QPR = 120^\circ$.

(a) Cosine rule: $QR^2 = PQ^2 + PR^2 - 2(PQ)(PR)\cos 120^\circ$ $= 100 + 49 - 2(10)(7)\left(-\frac{1}{2}\right)$ [M1] $= 149 + 70 = 219$ $QR = \sqrt{219}$ cm [A1]

(b) Sine rule: $\frac{\sin(\angle PQR)}{PR} = \frac{\sin 120^\circ}{QR}$ $\sin(\angle PQR) = \frac{7 \times \sin 120^\circ}{\sqrt{219}} = \frac{7 \times \frac{\sqrt{3}}{2}}{\sqrt{219}}$ [M1] $= \frac{7\sqrt{3}}{2\sqrt{219}} \approx 0.4096$ $\angle PQR = \sin^{-1}(0.4096) \approx 24.2^\circ$ (1 d.p.) [A1] [4 marks]

15. $\tan A = \frac{3}{4}$, $\tan B = \frac{5}{12}$, $A, B$ acute. $\tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$ [M1] $= \frac{\frac{3}{4} + \frac{5}{12}}{1 - \left(\frac{3}{4}\right)\left(\frac{5}{12}\right)}$ [M1] $= \frac{\frac{9}{12} + \frac{5}{12}}{1 - \frac{15}{48}} = \frac{\frac{14}{12}}{\frac{33}{48}} = \frac{14}{12} \times \frac{48}{33} = \frac{14 \times 4}{33} = \frac{56}{33}$ [A1] Since $\tan(A + B) > 0$ and $A, B$ acute, $A + B$ is acute. $A + B = \tan^{-1}\left(\frac{56}{33}\right) \approx 59.5^\circ$ (1 d.p.) Alternatively, note $\tan 45^\circ = 1$ and $\frac{56}{33} > 1$, so $A + B > 45^\circ$. Exact: $A + B = \tan^{-1}\left(\frac{56}{33}\right)$ [A1] [4 marks]

16. Prove $\displaystyle \frac{1 - \cos 2x}{\sin 2x} = \tan x$. LHS $= \frac{1 - (1 - 2\sin^2 x)}{2\sin x \cos x}$ [M1: use double angle formulas] $= \frac{2\sin^2 x}{2\sin x \cos x} = \frac{\sin x}{\cos x} = \tan x =$ RHS [A1] [2 marks]

17. $f(x) = \sin x + \cos x$, $0 \le x \le 2\pi$.

(a) $R = \sqrt{1^2 + 1^2} = \sqrt{2}$ [M1] $\tan \alpha = \frac{1}{1} = 1$, so $\alpha = \frac{\pi}{4}$ $\therefore f(x) = \sqrt{2}\sin\left(x + \frac{\pi}{4}\right)$ [A1]

(b) $f(x) = 1$: $\sqrt{2}\sin\left(x + \frac{\pi}{4}\right) = 1$ $\sin\left(x + \frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}$ [M1] $x + \frac{\pi}{4} = \frac{\pi}{4} + 2k\pi$ or $x + \frac{\pi}{4} = \frac{3\pi}{4} + 2k\pi$ $x = 0 + 2k\pi$ or $x = \frac{\pi}{2} + 2k\pi$ In $0 \le x \le 2\pi$: $x = 0, \frac{\pi}{2}, 2\pi$ [A1] [4 marks]

18. Triangle $XYZ$: $XY = 5$, $YZ = 8$, $\angle XYZ = 30^\circ$. Area $= \frac{1}{2}(XY)(YZ)\sin 30^\circ$ [M1] $= \frac{1}{2}(5)(8)\left(\frac{1}{2}\right) = 10$ cm$^2$ [A1] [2 marks]

19. $2\sin^2 x + 3\cos x = 0$, $0^\circ \le x \le 360^\circ$. Using $\sin^2 x = 1 - \cos^2 x$: $2(1 - \cos^2 x) + 3\cos x = 0$ [M1] $2 - 2\cos^2 x + 3\cos x = 0$ $2\cos^2 x - 3\cos x - 2 = 0$ [M1] $(2\cos x + 1)(\cos x - 2) = 0$ $\cos x = -\frac{1}{2}$ or $\cos x = 2$ (reject, since $-1 \le \cos x \le 1$) [M1] $\cos x = -\frac{1}{2}$: $x = 120^\circ, 240^\circ$ [A1] [4 marks]

20. Prove $\displaystyle \frac{\sin 3A}{\sin A} - \frac{\cos 3A}{\cos A} = 2$.

LHS $= \frac{\sin 3A \cos A - \cos 3A \sin A}{\sin A \cos A}$ [M1] $= \frac{\sin(3A - A)}{\sin A \cos A}$ [M1: using $\sin(P - Q) = \sin P \cos Q - \cos P \sin Q$] $= \frac{\sin 2A}{\sin A \cos A}$ [M1] $= \frac{2\sin A \cos A}{\sin A \cos A} = 2 =$ RHS [A1] [4 marks]

END OF ANSWER KEY