Graphs of Trigonometric Functions: Precalculus Math Worksheet Answer Key – PDF Download

Precalculus Trig Graphs Worksheet Answer Key: problems 1–5

Problem 1: Amplitude of a wave

The amplitude is \(|A|\).

1. 4
2. 3 (the sign only reflects the graph)
3. \(\frac{1}{2}\)
4. 5 (the +2 shifts the midline)

Problem 2: Period from B

Use period \(=\frac{2\pi}{|B|}\).

1. \(\frac{2\pi}{2}=\)\(\pi\)
2. \(\frac{2\pi}{1/2}=\)\(4\pi\)
3. \(\frac{2\pi}{3}\)
4. \(\frac{2\pi}{\pi}=\)\(2\)

Problem 3: Midline and range

The midline is \(y=D\) and the range is \(D-|A|\le y\le D+|A|\).

1. y=3; range \(1\le y\le 5\)
2. y=-1; range \(-5\le y\le 3\)
3. y=-2; range \(-2.5\le y\le -1.5\)
4. y=0.5; range \(-2.5\le y\le 3.5\)

Problem 4: Five key points

Use \(\cos x=1,0,-1,0,1\) and \(\sin x=0,1,0,-1,0\).

1. 3, 1, -1, 1, 3
2. 0, -1, 0, 1, 0

Problem 5: Read a cosine graph

Maximum \(2\), minimum \(-4\). Amplitude \(=\frac{2-(-4)}{2}=3\); midline \(y=\frac{2+(-4)}{2}=-1\). The maximum at \(x=0\) and the minimum at \(x=\frac{\pi}{2}\) are half a period apart, so the period is \(\pi\) and \(B=\frac{2\pi}{\pi}=2\).

Amplitude 3, period \(\pi\), midline \(y=-1\), \(y=3\cos 2x-1\)

Precalculus Trig Graphs Worksheet Answer Key: problems 6–10

Problem 6: Graph a reflected sine

The amplitude is 2 and the sign reflects the sine. Key points: \((0,0)\), \((\frac{\pi}{2},-2)\), \((\pi,0)\), \((\frac{3\pi}{2},2)\), \((2\pi,0)\).

Problem 7: Describe transformations

1. Reflect across the x-axis, shrink vertically by a factor of \(\frac{1}{2}\), and shift up 3.
2. Shift left \(\frac{\pi}{2}\). (The result is the graph of \(y=\cos x\).)
3. Stretch horizontally by a factor of 4, so the period becomes \(4\pi\).

Problem 8: Phase shift

Factor out \(B\) first.

1. Right \(\frac{\pi}{3}\)
2. Left \(\frac{\pi}{4}\)
3. \(\sin(2x-\pi)=\sin(2(x-\frac{\pi}{2}))\): right \(\frac{\pi}{2}\)
4. \(\cos(3x+\frac{\pi}{2})=\cos(3(x+\frac{\pi}{6}))\): left \(\frac{\pi}{6}\)

Problem 9: Read a tangent graph

Neighboring asymptotes are at \(x=-\frac{\pi}{4}\) and \(x=\frac{\pi}{4}\), so the period is \(\frac{\pi}{2}\), and \(B=\frac{\pi}{\pi/2}=2\). The asymptotes are \(x=\frac{\pi}{4}+\frac{k\pi}{2}\). At \(x=\frac{\pi}{8}\), \(\tan\frac{\pi}{4}=1\), and the graph shows \(y=1\), so \(A=1\).

\(y=\tan 2x\)

Problem 10: Tangent period and asymptotes

Period \(=\frac{\pi}{|B|}\); asymptotes where \(Bx=\pm\frac{\pi}{2}\).

1. Period \(\frac{\pi}{3}\); \(3x=\pm\frac{\pi}{2}\) gives \(x=\pm\frac{\pi}{6}\)
2. Period \(2\pi\); \(\frac{x}{2}=\pm\frac{\pi}{2}\) gives \(x=\pm\pi\)

Precalculus Trig Graphs Worksheet Answer Key: problems 11–15

Problem 11: Write a sine from a graph

The graph rises through the midline at \(x=\frac{\pi}{4}\) where \(y=2\), reaches a maximum 5 at \(x=\frac{\pi}{2}\), and a minimum \(-1\) at \(x=\pi\). Midline \(D=\frac{5+(-1)}{2}=2\), \(A=3\). A quarter period is \(\frac{\pi}{2}-\frac{\pi}{4}=\frac{\pi}{4}\), so the period is \(\pi\) and \(B=2\). A sine starts its cycle on the midline going up, at \(C=\frac{\pi}{4}\).

\(y=3\sin(2(x-\frac{\pi}{4}))+2\) (also \(y=-3\cos 2x+2\))

Problem 12: Evaluate a sinusoid

1. \(3\sin\frac{\pi}{6}+1=3\cdot\frac{1}{2}+1=\)2.5
2. \(3\sin\frac{\pi}{2}+1=\)4
3. \(3\sin\pi+1=\)1
4. \(3\sin\frac{3\pi}{2}+1=3(-1)+1=\)-2

Problem 13: Write an equation

1. \(B=\frac{2\pi}{\pi}=2\). Starting on the midline going up: \(y=5\sin 2x-2\)
2. \(B=\frac{2\pi}{8\pi}=\frac{1}{4}\). Starting at a maximum: \(y=3\cos(\frac{x}{4})+1\)
3. \(B=\frac{2\pi}{6}=\frac{\pi}{3}\): \(y=2\sin(\frac{\pi x}{3})\)

Problem 14: Identify B from a graph

Maximum 2, minimum \(-2\), so \(|A|=2\). The graph goes down first, so \(A=-2\). It crosses the axis at \(0\), \(2\pi\), \(4\pi\), so the period is \(4\pi\) and \(B=\frac{2\pi}{4\pi}=\frac{1}{2}\).

\(y=-2\sin(\frac{x}{2})\); amplitude 2, period \(4\pi\)

Problem 15: Shifted tangent

Asymptotes: \(x-\frac{\pi}{4}=\pm\frac{\pi}{2}\) gives \(x=-\frac{\pi}{4}\) and \(x=\frac{3\pi}{4}\). Center: \(x=\frac{\pi}{4}\), \(y=2\tan 0+1=1\). At \(x=0\): \(2\tan(-\frac{\pi}{4})+1=-1\). At \(x=\frac{\pi}{2}\): \(2\tan\frac{\pi}{4}+1=3\).

Asymptotes \(x=-\frac{\pi}{4}\), \(x=\frac{3\pi}{4}\); center \((\frac{\pi}{4},1)\); points \((0,-1)\) and \((\frac{\pi}{2},3)\)

Precalculus Trig Graphs Worksheet Answer Key: problems 16–20

Problem 16: Domain of a tangent

The function is undefined when \(2x=\frac{\pi}{2}+k\pi\), that is, when \(x=\frac{\pi}{4}+\frac{k\pi}{2}\).

All real numbers except \(x=\frac{\pi}{4}+\frac{k\pi}{2}\), \(k\) an integer

Problem 17: Ferris wheel height

1. Amplitude 40 ft; period \(\frac{2\pi}{\pi/2}=4\), so 4 minutes; midline \(h=50\)
2. \(h(1)=-40\cos\frac{\pi}{2}+50=\)50 ft; \(h(2)=-40\cos\pi+50=40+50=\)90 ft
3. \(h(0.5)=-40\cos\frac{\pi}{4}+50\approx -28.28+50=\)21.7 ft

Problem 18: Tide model

1. \(A=\frac{11-3}{2}=4\), \(D=\frac{11+3}{2}=7\). Half a period is \(9-3=6\), so the period is 12 and \(B=\frac{2\pi}{12}=\frac{\pi}{6}\).
2. \(h(t)=4\cos(\frac{\pi}{6}(t-3))+7\)
3. \(h(1)=4\cos(-\frac{\pi}{3})+7=4\cdot\frac{1}{2}+7=\)9 ft; \(h(6)=4\cos\frac{\pi}{2}+7=\)7 ft

Problem 19: Monthly temperature

\(A=\frac{74-30}{2}=22\), \(D=\frac{74+30}{2}=52\), period 12 months so \(B=\frac{\pi}{6}\). The maximum is at \(m=7\).

\(T(m)=22\cos(\frac{\pi}{6}(m-7))+52\)

\(T(4)=22\cos(-\frac{\pi}{2})+52=\)\(52^{\circ}\text{F}\); \(T(9)=22\cos\frac{\pi}{3}+52=11+52=\)\(63^{\circ}\text{F}\)

Problem 20: Bobbing buoy

\(A=1.5\), \(B=\frac{2\pi}{5}\), starting on the midline going up, so a sine: \(y(t)=1.5\sin(\frac{2\pi t}{5})\).

\(y(1.25)=1.5\sin\frac{\pi}{2}=\)1.5 ft; \(y(2.5)=1.5\sin\pi=\)0 ft; \(y(0.625)=1.5\sin\frac{\pi}{4}\approx 1.06\approx\)1.1 ft.

Precalculus Trig Graphs Worksheet Answer Key: problems 21–22

Problem 21: Hours of daylight

1. Amplitude 4 hours, midline \(D=12\), period 365 days. The longest day is day 172: \(D=16\).
2. \(D(80)=4\cos(\frac{2\pi}{365}(-92))+12\approx 4(-0.013)+12\approx\)11.9 hours; \(D(1)=4\cos(\frac{2\pi}{365}(-171))+12\approx 4(-0.981)+12\approx\)8.1 hours

Problem 22: Lighthouse beam

1. \(d=300\cdot\frac{1}{\sqrt{3}}\approx\)173.2 ft; \(300\cdot 1=\)300 ft; \(300\sqrt{3}\approx\)519.6 ft
2. \(d=300\tan 1.5\approx 300(14.10)\approx\)4,230.4 ft. The line \(\theta=\frac{\pi}{2}\) is a vertical asymptote: the beam becomes parallel to the shore and \(d\) grows without bound.

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