Trigonometric Functions and the Unit Circle: practice solutions, Grade 11 – download the PDF

1 Degrees to radians ★★★

Multiply each measure by \(\dfrac{\pi}{180}\).

1. \(45\times\dfrac{\pi}{180}=\dfrac{\pi}{4}\).
2. \(120\times\dfrac{\pi}{180}=\dfrac{2\pi}{3}\).
3. \(270\times\dfrac{\pi}{180}=\dfrac{3\pi}{2}\).
4. \(150\times\dfrac{\pi}{180}=\dfrac{5\pi}{6}\).

2 Radians to degrees ★★★

Multiply each measure by \(\dfrac{180}{\pi}\).

1. \(\dfrac{5\pi}{6}\times\dfrac{180}{\pi}=150^\circ\).
2. \(\dfrac{7\pi}{4}\times\dfrac{180}{\pi}=315^\circ\).
3. \(\dfrac{\pi}{9}\times\dfrac{180}{\pi}=20^\circ\).
4. \(\dfrac{11\pi}{6}\times\dfrac{180}{\pi}=330^\circ\).

3 Length of a sprinkler arc ★★★

Use \(s=r\theta\) with \(\theta\) in radians.

\(s=8\times\dfrac{3\pi}{4}=6\pi\approx 18.85\) inches.

In centimeters: \(18.85\times 2.54\approx 47.88\) cm.

Answer: the arc is about 18.85 inches long, or about 47.9 centimeters.

4 Special values on the unit circle ★★★

These come from the first-quadrant row of the exact values table.

1. \(\sin\dfrac{\pi}{6}=\dfrac{1}{2}\).
2. \(\cos\dfrac{\pi}{3}=\dfrac{1}{2}\).
3. \(\sin\dfrac{\pi}{4}=\dfrac{\sqrt{2}}{2}\).
4. \(\cos\dfrac{\pi}{2}=0\).
5. \(\tan\dfrac{\pi}{4}=\dfrac{\sqrt{2}/2}{\sqrt{2}/2}=1\).

5 Quadrants and signs ★★★

1. \(200^\circ\) is between \(180^\circ\) and \(270^\circ\): quadrant III, sine negative, cosine negative.
2. \(\dfrac{5\pi}{6}\) is between \(\dfrac{\pi}{2}\) and \(\pi\): quadrant II, sine positive, cosine negative.
3. \(300^\circ\) is between \(270^\circ\) and \(360^\circ\): quadrant IV, sine negative, cosine positive.
4. \(-\dfrac{\pi}{4}\) is a quarter of \(\pi\) clockwise, so it ends in quadrant IV: sine negative, cosine positive.

6 Amplitude and period ★★★

1. Amplitude \(|4|=4\); period \(2\pi\) because \(B=1\).
2. Amplitude 1; period \(\dfrac{2\pi}{3}\) because \(B=3\).
3. Amplitude \(|-2|=2\); \(B=\dfrac{1}{2}\), so the period is \(\dfrac{2\pi}{1/2}=4\pi\).

7 True or false? ★★★

1. True. The point at angle \(\pi\) on the unit circle is \((-1,0)\), whose x-coordinate is \(-1\).
2. False. At \(\dfrac{\pi}{2}\) the point is \((0,1)\), and \(\tan\dfrac{\pi}{2}=\dfrac{1}{0}\) is undefined. (The tangent equals 0 at \(0\) and \(\pi\).)
3. False. A straight angle is \(180^\circ=\pi\) radians; \(2\pi\) is a full turn.
4. True. This is the Pythagorean identity, valid for every angle, in degrees or radians.

8 Exact values with reference angles ★★★

1. Quadrant II, reference angle \(180^\circ-150^\circ=30^\circ\), sine positive: \(\sin150^\circ=\dfrac{1}{2}\).
2. Quadrant III, reference angle \(45^\circ\), cosine negative: \(\cos225^\circ=-\dfrac{\sqrt{2}}{2}\).
3. Quadrant IV, reference angle \(60^\circ\), tangent negative: \(\tan300^\circ=-\sqrt{3}\).
4. \(\dfrac{4\pi}{3}\) is in quadrant III with reference angle \(\dfrac{\pi}{3}\), sine negative: \(\sin\dfrac{4\pi}{3}=-\dfrac{\sqrt{3}}{2}\).

9 Cosine known, quadrant IV ★★★

By the Pythagorean identity, \(\sin^2\theta=1-\dfrac{9}{25}=\dfrac{16}{25}\), so \(\sin\theta=\pm\dfrac{4}{5}\).

Sine is negative in quadrant IV, so \(\sin\theta=-\dfrac{4}{5}\).

Then \(\tan\theta=\dfrac{-4/5}{3/5}=-\dfrac{4}{3}\).

10 Sine known, quadrant II ★★★

\(\cos^2\theta=1-\dfrac{25}{169}=\dfrac{144}{169}\), so \(\cos\theta=\pm\dfrac{12}{13}\).

Cosine is negative in quadrant II, so \(\cos\theta=-\dfrac{12}{13}\).

Then \(\tan\theta=\dfrac{5/13}{-12/13}=-\dfrac{5}{12}\).

11 Key points of a reflected cosine ★★★

1. Amplitude \(|-3|=3\); period \(\dfrac{2\pi}{2}=\pi\); midline \(y=1\).
2. Maximum \(1+3=4\); minimum \(1-3=-2\).
3. At \(x=0\): \(-3(1)+1=-2\). At \(\dfrac{\pi}{4}\): \(-3\cos\dfrac{\pi}{2}+1=1\). At \(\dfrac{\pi}{2}\): \(-3\cos\pi+1=4\). At \(\dfrac{3\pi}{4}\): \(-3\cos\dfrac{3\pi}{2}+1=1\). At \(\pi\): \(-3\cos2\pi+1=-2\).

The graph starts at its minimum because of the reflection (the negative sign).

12 A sine equation ★★★

Isolate: \(\sin x=\dfrac{1}{2}\). The reference angle is \(\dfrac{\pi}{6}\).

Sine is positive in quadrants I and II, so \(x=\dfrac{\pi}{6}\) or \(x=\pi-\dfrac{\pi}{6}=\dfrac{5\pi}{6}\).

13 A cosine equation ★★★

Isolate: \(\cos x=-\dfrac{\sqrt{2}}{2}\). The reference angle is \(\dfrac{\pi}{4}\).

Cosine is negative in quadrants II and III, so \(x=\pi-\dfrac{\pi}{4}=\dfrac{3\pi}{4}\) or \(x=\pi+\dfrac{\pi}{4}=\dfrac{5\pi}{4}\).

14 The Ferris wheel ★★★

1. Amplitude 20 and midline 25, so the lowest height is \(25-20=5\) ft and the highest is \(25+20=45\) ft. The period is \(\dfrac{2\pi}{\pi/6}=12\) minutes. In meters, \(45\times0.3048\approx13.7\) m.
2. \(h(2)=25-20\cos\dfrac{\pi}{3}=25-20\times\dfrac{1}{2}=15\) ft.
3. Solve \(25-20\cos\dfrac{\pi t}{6}=35\): \(\cos\dfrac{\pi t}{6}=-\dfrac{1}{2}\). For \(0\le t<12\) the angle \(\dfrac{\pi t}{6}\) is in \([0,2\pi)\), so \(\dfrac{\pi t}{6}=\dfrac{2\pi}{3}\) or \(\dfrac{4\pi}{3}\), giving \(t=4\) or \(t=8\).

Answer: the seat is at 35 feet after 4 minutes and after 8 minutes.

15 Harbor tide ★★★

1. Period \(\dfrac{2\pi}{\pi/6}=12\) hours; maximum \(8+3=11\) ft (about 3.35 m); minimum \(8-3=5\) ft.
2. \(d(1)=8+3\sin\dfrac{\pi}{6}=8+1.5=9.5\) ft.
3. Solve \(3\sin\dfrac{\pi t}{6}=1.5\), so \(\sin\dfrac{\pi t}{6}=\dfrac{1}{2}\). Then \(\dfrac{\pi t}{6}=\dfrac{\pi}{6}\) or \(\dfrac{5\pi}{6}\), giving \(t=1\) or \(t=5\).

Answer: the depth is 9.5 ft at 1 a.m. and at 5 a.m.

16 A shifted sine wave ★★★

1. Amplitude 2, period \(2\pi\), phase shift \(\dfrac{\pi}{3}\) to the right, range \([-2,2]\).
2. \(\sin\left(x-\dfrac{\pi}{3}\right)=0\) when \(x-\dfrac{\pi}{3}=0\) or \(\pi\), so \(x=\dfrac{\pi}{3}\) or \(x=\dfrac{4\pi}{3}\).
3. The maximum 2 occurs when \(x-\dfrac{\pi}{3}=\dfrac{\pi}{2}\), so at \(x=\dfrac{5\pi}{6}\). The minimum \(-2\) occurs when \(x-\dfrac{\pi}{3}=\dfrac{3\pi}{2}\), so at \(x=\dfrac{11\pi}{6}\).

17 One branch of a tangent ★★★

1. The period is \(\dfrac{\pi}{2}\).
2. \(\tan(2x)\) is undefined when \(2x=\pm\dfrac{\pi}{2}\), so the asymptotes are \(x=-\dfrac{\pi}{4}\) and \(x=\dfrac{\pi}{4}\).
3. \(\tan(2x)=0\) at \(x=0\); \(\tan(2x)=1\) when \(2x=\dfrac{\pi}{4}\), so \(x=\dfrac{\pi}{8}\); \(\tan(2x)=-1\) when \(2x=-\dfrac{\pi}{4}\), so \(x=-\dfrac{\pi}{8}\). The points are \(\left(-\dfrac{\pi}{8},-1\right)\), \((0,0)\) and \(\left(\dfrac{\pi}{8},1\right)\).

18 Factoring a trigonometric equation ★★★

Let \(u=\cos x\). Then \(2u^2-u-1=0\), which factors as \((2u+1)(u-1)=0\).

So \(\cos x=-\dfrac{1}{2}\) or \(\cos x=1\).

\(\cos x=-\dfrac{1}{2}\): reference angle \(\dfrac{\pi}{3}\), quadrants II and III, so \(x=\dfrac{2\pi}{3}\) or \(\dfrac{4\pi}{3}\).

\(\cos x=1\): \(x=0\).

Solutions: \(x=0,\ \dfrac{2\pi}{3},\ \dfrac{4\pi}{3}\).

19 Using the Pythagorean identity to solve ★★★

Replace \(\sin^2x\) by \(1-\cos^2x\): \(2-2\cos^2x+3\cos x=3\), which gives \(2\cos^2x-3\cos x+1=0\).

Factor: \((2\cos x-1)(\cos x-1)=0\), so \(\cos x=\dfrac{1}{2}\) or \(\cos x=1\).

\(\cos x=\dfrac{1}{2}\): \(x=\dfrac{\pi}{3}\) or \(\dfrac{5\pi}{3}\). \(\cos x=1\): \(x=0\).

Solutions: \(x=0,\ \dfrac{\pi}{3},\ \dfrac{5\pi}{3}\).

20 Writing an equation from features ★★★

1. Amplitude \(\dfrac{5-(-3)}{2}=4\); midline \(\dfrac{5+(-3)}{2}=1\).
2. \(B=\dfrac{2\pi}{\pi}=2\). A maximum at \(x=0\) means an unreflected cosine, so \(y=4\cos(2x)+1\).
3. At \(x=0\): \(4\cos0+1=5\), the maximum. At \(x=\dfrac{\pi}{2}\): \(4\cos\pi+1=-3\), the minimum, half a period later.

21 Daily temperature ★★★

1. The cosine equals 1 at \(t=4\), giving the minimum \(62-14=48^\circ\)F at 4 a.m.; it equals \(-1\) at \(t=16\), giving the maximum \(62+14=76^\circ\)F at 4 p.m.
2. \(T(10)=62-14\cos\dfrac{\pi}{2}=62^\circ\)F. \(T(12)=62-14\cos\dfrac{2\pi}{3}=62-14\times\left(-\dfrac{1}{2}\right)=69^\circ\)F.
3. Solve \(62-14\cos u=69\) with \(u=\dfrac{\pi(t-4)}{12}\): \(\cos u=-\dfrac{1}{2}\). For \(0\le t<24\), \(u\) runs from \(-\dfrac{\pi}{3}\) up to just below \(\dfrac{5\pi}{3}\), so \(u=\dfrac{2\pi}{3}\) or \(\dfrac{4\pi}{3}\). Then \(t-4=8\) or \(16\), so \(t=12\) or \(t=20\). In Celsius, \(\dfrac{5}{9}(69-32)\approx20.6^\circ\)C.

Answer: 69°F (about 20.6°C) occurs at noon and at 8 p.m.

22 Proving and checking an identity ★★★

(a) By the Pythagorean identity, \(\sin^2x=1-\cos^2x=(1-\cos x)(1+\cos x)\). Dividing by \(1-\cos x\) (not zero) gives \(\dfrac{\sin^2x}{1-\cos x}=1+\cos x\).

(b) At \(x=\dfrac{\pi}{3}\): \(\sin^2x=\dfrac{3}{4}\) and \(1-\cos x=\dfrac{1}{2}\), so the left side is \(\dfrac{3/4}{1/2}=\dfrac{3}{2}\). The right side is \(1+\dfrac{1}{2}=\dfrac{3}{2}\). Both sides agree.