Graphs of Trigonometric Functions: practice solutions, Grade 12 – download the PDF

1 Amplitude and period ★★★

In \(y=A\sin(Bx)\), the amplitude is \(|A|=4\) and the period is \(\dfrac{2\pi}{|B|}=\dfrac{2\pi}{3}\).

The maximum is \(4\) and the minimum is \(-4\) because there is no vertical shift.

2 A reflected cosine ★★★

The amplitude is \(|-2|=2\). Here \(B=\dfrac12\), so the period is \(\dfrac{2\pi}{1/2}=4\pi\).

Because of the minus sign, the graph is reflected: at \(x=0\) we get \(y=-2\), a minimum. The maximum \(+2\) occurs half a period later, at \(x=2\pi\) (check: \(\cos\pi=-1\), so \(y=-2\cdot(-1)=2\)).

3 Midline, maximum, minimum ★★★

The midline is \(y=5\). The amplitude is 2, so the maximum is \(5+2=7\) and the minimum is \(5-2=3\).

Sine is largest at \(x=\dfrac{\pi}{2}\) (value 7) and smallest at \(x=\dfrac{3\pi}{2}\) (value 3).

4 A table of values ★★★

Use \(\cos 0=1\), \(\cos\dfrac{\pi}{3}=\dfrac12\), \(\cos\dfrac{\pi}{2}=0\), \(\cos\dfrac{2\pi}{3}=-\dfrac12\), \(\cos\pi=-1\) and multiply by 3:

The points describe half a cycle of a cosine wave going from its maximum 3 down to its minimum \(-3\).

5 True or false? ★★★

1. True: \(\cos(x+2\pi)=\cos x\) and no smaller positive number works.
2. False: tangent takes every real value, for example \(\tan\dfrac{\pi}{4}=1\) and \(\tan\dfrac{\pi}{3}=\sqrt3\approx1.73\).
3. False: \(\sec x=\dfrac{1}{\cos x}\) is undefined when \(\cos x=0\), for example at \(x=\dfrac{\pi}{2}\).
4. True: sine is odd, \(\sin(-x)=-\sin x\), so the graph is symmetric about the origin.
5. False: an amplitude is a distance and is never negative; it is \(|-4|=4\) (the minus sign only reflects the graph).

6 Asymptotes of a tangent ★★★

The period is \(\dfrac{\pi}{|B|}=\dfrac{\pi}{2}\).

Asymptotes occur where \(2x=\dfrac{\pi}{2}+k\pi\), i.e. \(x=\dfrac{\pi}{4}+\dfrac{k\pi}{2}\). Taking \(k=0\) gives \(\dfrac{\pi}{4}\); \(k=-1\) gives \(-\dfrac{\pi}{4}\); \(k=1\) gives \(\dfrac{3\pi}{4}\) and \(k=-2\) gives \(-\dfrac{3\pi}{4}\), both outside the interval. So the asymptotes are \(x=-\dfrac{\pi}{4}\) and \(x=\dfrac{\pi}{4}\).

7 Inverse values ★★★

1. The angle in \(\left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right]\) with sine \(\dfrac12\) is \(\dfrac{\pi}{6}\).
2. The angle in \([0,\pi]\) with cosine \(-\dfrac12\) is \(\dfrac{2\pi}{3}\).
3. \(\tan\dfrac{\pi}{4}=1\), so \(\arctan1=\dfrac{\pi}{4}\).
4. \(\sin\left(-\dfrac{\pi}{4}\right)=-\dfrac{\sqrt2}{2}\) and \(-\dfrac{\pi}{4}\) lies in the range, so the answer is \(-\dfrac{\pi}{4}\).

8 Cosine becomes sine ★★★

Writing \(x+\dfrac{\pi}{2}=x-\left(-\dfrac{\pi}{2}\right)\), the phase shift is \(-\dfrac{\pi}{2}\): the graph of sine is slid \(\dfrac{\pi}{2}\) units to the left.

The result starts at its maximum 1 when \(x=0\), so it is \(y=\cos x\). For instance, at \(x=\dfrac{\pi}{3}\): \(\sin\dfrac{5\pi}{6}=\dfrac12=\cos\dfrac{\pi}{3}\).

9 Factor before you read ★★★

Factor: \(3x+\pi=3\left(x+\dfrac{\pi}{3}\right)\), so \(y=-4\cos\left(3\left(x+\dfrac{\pi}{3}\right)\right)+2\).

Amplitude \(4\); period \(\dfrac{2\pi}{3}\); phase shift \(\dfrac{\pi}{3}\) to the left; midline \(y=2\); maximum \(2+4=6\); minimum \(2-4=-2\).

At \(x=0\): \(y=-4\cos\pi+2=-4(-1)+2=6\), the maximum, which is consistent.

10 Equation of a sine wave ★★★

\(A=6\), \(D=2\), \(B=\dfrac{2\pi}{10}=\dfrac{\pi}{5}\). The wave starts on the midline going up, so \(y=6\sin\left(\dfrac{\pi x}{5}\right)+2\).

At \(x=2.5\): \(\sin\dfrac{\pi}{2}=1\), so \(y=8\) (maximum). At \(x=7.5\): \(\sin\dfrac{3\pi}{2}=-1\), so \(y=-4\) (minimum).

11 A cosine from its extremes ★★★

\(A=\dfrac{7-(-1)}{2}=4\) and \(D=\dfrac{7+(-1)}{2}=3\). The period is \(\dfrac{5\pi}{4}-\dfrac{\pi}{4}=\pi\), so \(B=2\). A maximum at \(x=\dfrac{\pi}{4}\) means \(C=\dfrac{\pi}{4}\).

\(y=4\cos\left(2\left(x-\dfrac{\pi}{4}\right)\right)+3\). Check: at \(x=\dfrac{\pi}{4}\), \(y=4+3=7\); at \(x=\dfrac{3\pi}{4}\), \(y=4\cos\pi+3=-1\).

12 Solving trigonometric equations ★★★

1. Reference angle \(\dfrac{\pi}{6}\), sine positive in quadrants I and II: \(x=\dfrac{\pi}{6}\) or \(x=\dfrac{5\pi}{6}\).
2. \(\cos x=-\dfrac{\sqrt3}{2}\): reference angle \(\dfrac{\pi}{6}\), cosine negative in quadrants II and III: \(x=\dfrac{5\pi}{6}\) or \(\dfrac{7\pi}{6}\).
3. Reference angle \(\dfrac{\pi}{4}\), tangent negative in quadrants II and IV: \(x=\dfrac{3\pi}{4}\) or \(\dfrac{7\pi}{4}\).

13 A shifted tangent ★★★

Period: \(\dfrac{\pi}{1/2}=2\pi\). Asymptotes: \(\dfrac{x}{2}=\dfrac{\pi}{2}+k\pi\), i.e. \(x=\pi+2k\pi\); in \((-2\pi,2\pi)\) these are \(x=-\pi\) and \(x=\pi\).

At \(x=0\): \(y=0+1=1\). At \(x=\dfrac{\pi}{2}\): \(y=\tan\dfrac{\pi}{4}+1=2\).

Intercept: \(\tan\dfrac{x}{2}=-1\Rightarrow\dfrac{x}{2}=-\dfrac{\pi}{4}\Rightarrow x=-\dfrac{\pi}{2}\).

14 A stretched secant ★★★

(a) \(y=\dfrac{2}{\cos x}\) is defined when \(\cos x\neq0\): \(x\neq\dfrac{\pi}{2}+k\pi\). Since \(|\cos x|\le1\), \(|y|\ge2\): the range is \((-\infty,-2]\cup[2,\infty)\).

(b) \(y\left(\dfrac{\pi}{3}\right)=\dfrac{2}{1/2}=4\) and \(y\left(\dfrac{2\pi}{3}\right)=\dfrac{2}{-1/2}=-4\).

(c) At \(x=0\), \(\cos x=1\) so \(y=2\), a local minimum. At \(x=\pi\), \(\cos x=-1\) so \(y=-2\), a local maximum.

15 Compositions with inverses ★★★

1. 0.3 lies in \([-1,1]\), so the answer is \(0.3\).
2. Right triangle with opposite 8 and hypotenuse 17: adjacent \(=\sqrt{17^2-8^2}=15\). So \(\cos=\dfrac{15}{17}\).
3. Adjacent 5, hypotenuse 13: opposite \(=12\). So \(\tan=\dfrac{12}{5}\).
4. Opposite 3, adjacent 4: hypotenuse \(5\). So \(\sin=\dfrac35\).

16 A bicycle pedal ★★★

(a) Maximum \(12+7=19\) in (about 48.3 cm); minimum \(12-7=5\) in (about 12.7 cm).

(b) Period \(=\dfrac{2\pi}{2.5\pi}=0.8\) s. In a minute: \(\dfrac{60}{0.8}=75\) revolutions per minute.

(c) \(h(0.1)=12+7\sin(0.25\pi)=12+7\cdot\dfrac{\sqrt2}{2}\approx16.95\) in.

17 A sound wave ★★★

Amplitude: 0.4. Here \(B=2\pi\cdot262\), so the period is \(\dfrac{2\pi}{2\pi\cdot262}=\dfrac{1}{262}\approx0.003817\) s, about 3.82 ms.

In 0.05 s there are \(0.05\cdot262=13.1\) cycles, so 13 full cycles.

18 Tide model ★★★

(a) Amplitude 2.5 ft; period \(\dfrac{2\pi}{\pi/6}=12\) h; maximum \(8.5\) ft; minimum \(3.5\) ft.

(b) \(d(2)=6+2.5\cos\dfrac{\pi}{3}=6+1.25=7.25\) ft.

(c) Solve \(6+2.5\cos\dfrac{\pi t}{6}\ge7.75\Leftrightarrow\cos\dfrac{\pi t}{6}\ge0.7\). Around \(t=0\), \(\left|\dfrac{\pi t}{6}\right|\le\arccos0.7\approx0.7954\), so \(|t|\le\dfrac{6}{\pi}(0.7954)\approx1.519\). The boat can float for about \(2\times1.519\approx3.04\) hours.

19 Reading an equation from a graph ★★★

(a) Maximum 4, minimum \(-2\): amplitude \(\dfrac{4-(-2)}{2}=3\), midline \(y=\dfrac{4+(-2)}{2}=1\). The curve repeats from \(x=0\) to \(x=8\): period 8.

(b) \(B=\dfrac{2\pi}{8}=\dfrac{\pi}{4}\). The graph starts at a minimum, so use a negative cosine: \(y=-3\cos\left(\dfrac{\pi x}{4}\right)+1\). Check: \(y(0)=-2\), \(y(2)=1\), \(y(4)=4\).

(c) The rising midline crossing is at \(x=2\), so \(C=2\): \(y=3\sin\left(\dfrac{\pi}{4}(x-2)\right)+1\). Check: \(y(2)=1\), \(y(4)=3\sin\dfrac{\pi}{2}+1=4\).

20 Inverse of the function, not the identity ★★★

1. \(\sin\dfrac{4\pi}{3}=-\dfrac{\sqrt3}{2}\). The angle in \(\left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right]\) with that sine is \(-\dfrac{\pi}{3}\). (\(\dfrac{4\pi}{3}\) is outside the range.)
2. \(\cos\left(-\dfrac{\pi}{4}\right)=\dfrac{\sqrt2}{2}\), and the angle in \([0,\pi]\) is \(\dfrac{\pi}{4}\). (\(-\dfrac{\pi}{4}\) is outside \([0,\pi]\).)
3. \(\cos\dfrac{7\pi}{6}=-\dfrac{\sqrt3}{2}\), so the answer is \(\dfrac{5\pi}{6}\) in \([0,\pi]\).
4. \(\tan\dfrac{3\pi}{4}=-1\), so the answer is \(-\dfrac{\pi}{4}\) in \(\left(-\dfrac{\pi}{2},\dfrac{\pi}{2}\right)\).

In every case the input angle was outside the restricted range of the inverse, so the inverse returns the equivalent angle inside it.

21 An identity for inverse functions ★★★

Let \(\theta=\arcsin x\), so \(\theta\in\left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right]\) and \(\sin\theta=x\). Then \(\dfrac{\pi}{2}-\theta\in[0,\pi]\) and \(\cos\left(\dfrac{\pi}{2}-\theta\right)=\sin\theta=x\).

The only angle in \([0,\pi]\) whose cosine is \(x\) is \(\arccos x\), so \(\arccos x=\dfrac{\pi}{2}-\theta\), which gives \(\arcsin x+\arccos x=\dfrac{\pi}{2}\).

Check: \(\arcsin0.6\approx0.6435\) and \(\arccos0.6\approx0.9273\); their sum is \(1.5708\approx\dfrac{\pi}{2}\).

22 A decreasing tangent ★★★

(a) Period \(\dfrac{\pi}{\pi/4}=4\).

(b) Asymptotes where \(\dfrac{\pi x}{4}=\dfrac{\pi}{2}+k\pi\Rightarrow x=2+4k\): in \((-4,4)\) they are \(x=-2\) and \(x=2\). The only intercept in \((-2,2)\) is \(x=0\).

(c) \(g(1)=-2\tan\dfrac{\pi}{4}=-2\) and \(g(-1)=-2\tan\left(-\dfrac{\pi}{4}\right)=2\).

(d) Tangent increases; multiplying by \(-2\) reverses it, so \(g\) is decreasing on each branch (it falls from \(+\infty\) to \(-\infty\)).

23 Daily temperature ★★★

(a) Maximum \(73+11=84\) \(^\circ\)F at \(t=15\) (3 PM); minimum \(73-11=62\) \(^\circ\)F at \(t=3\) (3 AM), since \(\cos(-\pi)=-1\).

(b) \(T(9)=73+11\cos\left(-\dfrac{\pi}{2}\right)=73\) \(^\circ\)F. \(T(18)=73+11\cos\dfrac{\pi}{4}\approx80.78\) \(^\circ\)F, which is \(\dfrac{80.78-32}{1.8}\approx27.1\) \(^\circ\)C.

(c) We need \(\cos\dfrac{\pi(t-15)}{12}\ge\dfrac{7}{11}\approx0.6364\), so \(|t-15|\le\dfrac{12}{\pi}\arccos\dfrac{7}{11}\approx\dfrac{12}{\pi}(0.8810)\approx3.365\) h. Hence \(11.63\le t\le18.37\): from about 11:38 AM to about 6:22 PM, roughly 6.7 hours.