Sec 4 A Math: Trigonometric Equations & Graphs, practice questions & worked solutions

(a)

$$-90^\circ<\tan^{-1}p<90^\circ \text{or} -\frac{\pi}{2}<\tan^{-1}p<\frac{\pi}{2}$$

(b)

$$\pi-x$$

$$\begin{aligned} 9\sin x &= -2\sec x\\ 9\sin x &= -\frac{2}{\cos x}\\ \sin x \cos x &= -\frac{2}{9}\\ 2\sin x \cos x &= -\frac{4}{9}\\ \sin 2x &= -\frac{4}{9} \end{aligned}$$ $$\begin{aligned} \text{Basic angle} &= \sin^{-1}\!\left(\frac{4}{9}\right)\\ &= 0.46055\\ 2x &= -0.46055,\ -(\pi - 0.46055)\\ &= -0.46055 \ \text{or} \ -2.68104\\ x &= -0.230 \ \text{or} \ -1.34 \end{aligned}$$

$$\begin{aligned} \frac{5\cos x + \sin 2x}{\cos 2x + 5\sin x} &= \cot x\\ \frac{5\cos x + \sin 2x}{\cos 2x + 5\sin x} &= \frac{\cos x}{\sin x}\\ \sin x\,(5\cos x + \sin 2x) &= \cos x\,(\cos 2x + 5\sin x)\\ \sin x\,(5\cos x + 2\sin x \cos x) &= \cos x\,(\cos^2 x - \sin^2 x + 5\sin x)\\ 5\sin x \cos x + 2\sin^2 x \cos x &= \cos^3 x - \sin^2 x \cos x + 5\sin x \cos x\\ 2\sin^2 x \cos x &= \cos^3 x - \sin^2 x \cos x\\ 3\sin^2 x \cos x - \cos^3 x &= 0\\ \cos x\,(3\sin^2 x - \cos^2 x) &= 0 \end{aligned}$$ $$\begin{array}{c|c} \cos x = 0 & 3\sin^2 x - \cos^2 x = 0\\[2pt] x = 90^\circ & 3\sin^2 x = \cos^2 x\\[2pt] & 3\tan^2 x = 1\\[2pt] & \tan^2 x = \dfrac{1}{3}\\[4pt] & \tan x = \pm\dfrac{1}{\sqrt{3}}\\[4pt] & x = 30^\circ,\ 150^\circ \end{array}$$ $$\therefore x = 30^\circ,\ 90^\circ,\ 150^\circ$$

(a)

$$h = \frac{2\pi}{\pi/6} = 12\ \text{hours}$$

(b)

$h$ is lowest when $\sin\!\left(\dfrac{\pi}{6}t\right)=-1$: $$\begin{aligned} \frac{\pi}{6}t &= \frac{3\pi}{2}\\ t &= 9 \end{aligned}$$ Time is $9$ am (or $0900$ hrs).

(c)

$$\begin{aligned} 7+2\sin\!\left(\tfrac{\pi}{6}t\right) &= 5.6\\ \sin\!\left(\tfrac{\pi}{6}t\right) &= -\frac{7}{10} \end{aligned}$$ basic angle $=\sin^{-1}\!\left(\dfrac{7}{10}\right)\approx 0.77539$ $$\begin{aligned} \frac{\pi}{6}t &= \pi+0.77539\ (\approx 3.9169),\ 2\pi-0.77539\ (\approx 5.5077)\\ t &= 7.4808,\ 10.519 \end{aligned}$$ Two critical timings are $7.29$ am (rejected since boat is still unloading) and $10.31$ am. Hence, earliest time boat can leave the port is $10.31$ am or $10.32$ am (or 24 hr format).

(a)

$$-\frac{\pi}{2}<\tan^{-1}x<\frac{\pi}{2}$$

(b)

(i) $$\begin{aligned} \text{period} &= \frac{2\pi}{c}=\pi\\ c &= 2\\ a-b &= 7\\ a+b &= 1\\ a &= 4\\ b &= -3\\ y &= 4-3\cos 2x \end{aligned}$$

(ii) $$\begin{aligned} a\pi+b\pi\cos cx &= 5\pi-2x\\ a+b\cos cx &= 5-\frac{2}{\pi}x \end{aligned}$$ There are $3$ solutions.

(a)

$h = 2(1 - \cos kt)$.

Since the diameter of the carousel $=$ max value of $h$, $h = 2(1 - (-1)) = 4$ m

(b)

Period $= 20$ s $$\begin{aligned} \frac{2\pi}{k} &= 20\\ k &= \frac{\pi}{10}\ rad/s \end{aligned}$$

Alternative method

$$\begin{aligned} &2\pi \text{ rad in 20 s}\\ &\frac{2\pi}{20} \text{ rad in 1 s}\\ &\frac{2\pi}{20}t \text{ rad in } t \text{ s}\\ k &= \frac{\pi}{10} \end{aligned}$$

(c)

$$\begin{aligned} 1 &= 2\left(1 - \cos\frac{\pi}{10}t\right)\\ \frac{1}{2} &= \cos\frac{\pi}{10}t \end{aligned}$$ $\text{basic } \angle = \dfrac{\pi}{3}$ (solution in 1st and 4th quad) $$\begin{aligned} \frac{\pi}{10}t &= \frac{\pi}{3},\ \frac{5\pi}{3}\\ t &= \frac{10}{3},\ \frac{50}{3} \end{aligned}$$ Not able to view: $\dfrac{50}{3} - \dfrac{10}{3} = \dfrac{40}{3}\ s$ or $13.3\ s$ (3 s.f.)

(a)

$$\begin{aligned} 2\cos 2x &= 4\sin x + 3\\ 2(1 - 2\sin^2 x) &= 4\sin x + 3\\ 2 - 4\sin^2 x - 4\sin x - 3 &= 0\\ 4\sin^2 x + 4\sin x + 1 &= 0\\ (2\sin x + 1)^2 &= 0\\ \sin x &= -\frac{1}{2}\\ \alpha = \frac{\pi}{6} x &= -\frac{\pi}{6},\ -\frac{5\pi}{6} \end{aligned}$$

(b)

$$\begin{aligned} 3\cosec^2 x - \cosec x &= 2 - \cot^2 x\\ 3\cosec^2 x - \cosec x &= 2 - (\cosec^2 x - 1)\\ 4\cosec^2 x - \cosec x - 3 &= 0\\ (4\cosec x + 3)(\cosec x - 1) &= 0 \end{aligned}$$ $$4\cosec x + 3 = 0 \qquad\qquad \cosec x - 1 = 0$$ $$\cosec x = -\frac{3}{4} \qquad\qquad \cosec x = 1$$ $$\sin x = -\frac{4}{3}\ (\text{rej}) \qquad\qquad \sin x = 1$$ $$x = 90^\circ$$

Alternate Solution

$$\begin{aligned} \frac{3}{\sin^2 x} - \frac{1}{\sin x} &= 2 - \frac{\cos^2 x}{\sin^2 x}\\ 3 - \sin x &= 2\sin^2 x - \cos^2 x\\ 3 - \sin x &= 2\sin^2 x - (1 - \sin^2 x)\\ 3\sin^2 x + \sin x - 4 &= 0\\ (3\sin x + 4)(\sin x - 1) &= 0\\ \sin x = -\frac{4}{3}\ (NA) &\text{or} \sin x = 1\\ x &= 90^\circ \end{aligned}$$

(a)

$$a = 3$$ Sub $\left(\dfrac{2\pi}{3}, 0\right)$ into $y = a + 6\cos bx$: $$\begin{aligned} 0 &= 3 + 6\cos b\!\left(\tfrac{2\pi}{3}\right)\\ \cos b\!\left(\tfrac{2\pi}{3}\right) &= -\frac{1}{2}\\ b\!\left(\tfrac{2\pi}{3}\right) &= \tfrac{2\pi}{3}\ (\text{rej}),\ \tfrac{4\pi}{3}\\ b &= 2 \end{aligned}$$

(b)

(i) $y = 1 - \dfrac{2x}{\pi} \implies y = -3\sin x$ at the points of intersection.

(ii) $$\begin{aligned} 2x &= \pi + 3\pi\sin(x)\\ \frac{2x}{\pi} &= 1 + 3\sin x\\ -3\sin(x) &= 1 - \frac{2x}{\pi} \end{aligned}$$ No. of solution(s) $= 1$

(a)

$$a = 2, b = 3, c = 3$$

(b)

(c)

$$\begin{aligned} \cos 3x &= d\\ 2\cos 3x &= 2d\\ 2\cos 3x + 3 &= 2d + 3\\ y &= 2d + 3 \end{aligned}$$ When $y = 1$, $$\begin{aligned} 2d + 3 &= 1\\ d &= -1 \end{aligned}$$

Or

min point $\cos 3x = d$ $$d = -1$$

(i)

$$\begin{aligned} 5\sin^2 x - \cos^2 x - 1 &= 5\left(\frac{1 - \cos 2x}{2}\right) - \left(\frac{\cos 2x + 1}{2}\right) - 1\\ &= \frac{5}{2} - \frac{5\cos 2x}{2} - \frac{1}{2}\cos 2x - \frac{1}{2} - 1\\ &= 1 - 3\cos 2x \end{aligned}$$

(ii)

$$5\sin^2 x - \cos^2 x - 1 = 1 - 3\cos 2x$$ $$\begin{aligned} \text{Amplitude} &= 3\\ \text{Period} &= \frac{2\pi}{2}\\ &= \pi \text{ radian} \end{aligned}$$

(iii)

(iv)

$$y = -\frac{3}{2\pi}x + 1$$ $$\therefore \text{Number of solutions} = 2.$$