Sec 3 A Math: Trigonometric Equations & Graphs, practice questions & worked solutions

(a)

$$\begin{aligned} \text{Period} &= 720^\circ \text{ or } 4\pi\\ \text{Amplitude} &= 1 \end{aligned}$$

(b)

$$\begin{aligned} \text{Period} &= 360^\circ \text{ or } 2\pi\\ \text{Amplitude} &= 2 \end{aligned}$$

(c)

$$\begin{aligned} 2\cot^2 2x &= 1-\frac{5}{\sin 2x}\\ 2\cot^2 2x &= 1-5\,\mathrm{cosec}\,2x\\ 2(\mathrm{cosec}^2\,2x-1) &= 1-5\,\mathrm{cosec}\,2x\\ 2\,\mathrm{cosec}^2\,2x+5\,\mathrm{cosec}\,2x-3 &= 0 \end{aligned}$$ Let $y=\mathrm{cosec}\,2x$. $$\begin{aligned} 2y^2+5y-3 &= 0\\ (2y-1)(y+3) &= 0\\ y=-3 \;&\text{or}\; y=\tfrac{1}{2}\\ \mathrm{cosec}\,2x=-3 \;&\text{or}\; \mathrm{cosec}\,2x=\tfrac{1}{2}\\ \sin 2x=-\tfrac{1}{3} \;&\text{or}\; \sin 2x=2\ \text{(rej.)} \end{aligned}$$ $$\begin{aligned} \alpha &= \sin^{-1}\tfrac{1}{3}\\ &= 19.471^\circ \end{aligned}$$ $$\begin{aligned} 2x &= 199.471^\circ,\ 340.529^\circ\\ x &= 99.7^\circ,\ 170.3^\circ (1\text{ d.p})\\[4pt] 2x &= -19.471^\circ,\ -160.529^\circ\\ x &= -9.7^\circ,\ -80.3^\circ (1\text{ d.p}) \end{aligned}$$

(a)

Amplitude $= 4$.

(b)

$$\begin{aligned} \text{Period} &= 4\pi\\ \frac{2\pi}{a} &= 4\pi\\ a &= \frac{1}{2} \end{aligned}$$ $$\begin{aligned} 4+b &= 6\\ b &= 2 \end{aligned}$$

(c)

$f(x)=4\cos\tfrac{1}{2}x+2$ for $0\le x\le 3\pi$.

(a)

$$\begin{aligned} \text{Period} &= \frac{\pi}{2}\\ \text{Period, } \frac{2\pi}{b} &= \frac{\pi}{2}\\ b &= 2\pi\div\frac{\pi}{2}=4 \end{aligned}$$

(b)

$y=c$ is the axis of curve $$\begin{aligned} c &= \frac{\text{maximum}+\text{minimum}}{2}\\ &= \frac{4+(-2)}{2}\\ c &= 1 \end{aligned}$$

(c)

By observation, $a<0$. $$\begin{gathered} a=-(4-1) \text{OR} a=-[1-(-2)]\\ \text{OR} a=-\!\left(\frac{\text{maximum}-\text{minimum}}{2}\right)=-\!\left(\frac{4-(-2)}{2}\right)=-3 \end{gathered}$$ Equation of curve is $y=-3\cos 4x+1$.

(a)

(b)

The $2\cos 3x=\dfrac{2}{\pi}x$ is equivalent to $-2\cos 3x+1=-\dfrac{2}{\pi}x+1$ and so its solutions can be found from the interception points of the two graphs.

(a)

$y = c$ is axis of curve $$\begin{aligned} c &= \frac{\text{maximum} + \text{minimum}}{2}\\ &= \frac{2 + (-8)}{2}\\ &= -3 \end{aligned}$$

(b)

$$\begin{aligned} \frac{2\pi}{b} &= \text{period}\\ \frac{2\pi}{b} &= \frac{\pi}{2}\\ b &= 4 \end{aligned}$$

(c)

$$y = a\sin bx + c$$ Since this is a negative sine graph and the amplitude is 5, $a = -5$.

The equation of the curve is $y = -5\sin 4x - 3$.

(a)

$$b = \frac{6 + 2}{2} = 4$$ $$k = \frac{2\pi}{12} = \frac{\pi}{6}$$

(b)

(c)

2 times

(a)

$$\begin{aligned} \text{Amplitude} &= 2\\ \text{Period} &= \frac{2\pi}{3}\ (\text{or } 120^\circ) \end{aligned}$$

(b)

$y=2\sin 3x+1$ for $0\le x\le\pi$:

(c)

$$\begin{aligned} \pi\sin 3x-x &= 0\\ 2\sin 3x-\frac{2x}{\pi} &= 0\\ 2\sin 3x+1 &= \frac{2x}{\pi}+1 \end{aligned}$$ Draw $y=\dfrac{2x}{\pi}+1$.

Number of solutions $=4$.

(a)

Amplitude $= 3$

Period $= 4\pi$ or $720^\circ$

(b)

(c)

$$\begin{aligned} 3\sin\frac{x}{2} + 1 &= -2\cos\frac{x}{2} \end{aligned}$$ Sketch $y = -2\cos\dfrac{x}{2}$

Number of solutions $= 2$

(a)

$$\begin{aligned} \text{Period: } \frac{2\pi}{\frac{\pi}{b}} &= 20\\ b &= 10 \end{aligned}$$ Max $h = 90$

Min $h = 10$ $$\begin{aligned} a + c &= 90\\ -a + c &= 10\\ 2c &= 100\\ c &= 50\\ a = 90 - 50 = 40 &\text{or} a = \frac{90 - 10}{2} = 40 \end{aligned}$$ $$\therefore h = 40\cos\left(\frac{\pi}{10}t\right) + 50$$

(b)

(d)

(c)

$$\begin{aligned} 62 &= 40\cos\left(\frac{\pi}{10}t\right) + 50\\ 40\cos\left(\frac{\pi}{10}t\right) &= 12\\ \cos\left(\frac{\pi}{10}t\right) &= \frac{12}{40}\\ \alpha &= 1.26610\\ \frac{\pi}{10}t &= 1.26610\\ t &= 4.03 \text{ s (3sf)} \end{aligned}$$

(d)

(ii) $$\begin{aligned} t &= 40\cos\left(\frac{\pi}{10}t\right) + 35\\ t + 15 &= 40\cos\left(\frac{\pi}{10}t\right) + 50 \end{aligned}$$ $3$ solutions