Trigonometric Identities: Precalculus Math Worksheet Answer Key – PDF Download

Precalculus Trig Identities Worksheet Answer Key: problems 1–5

Problem 1: Reciprocal and quotient forms

Apply the reciprocal and quotient identities.

1. \(\frac{1}{\cos\theta}\)
2. \(\frac{1}{\sin\theta}\)
3. \(\frac{\cos\theta}{\sin\theta}\)
4. \(\frac{\sin\theta}{\cos\theta}\cdot\frac{1}{\sin\theta}=\)\(\frac{1}{\cos\theta}\)

Problem 2: Reciprocal values

1. \(\text{csc}\,\theta=\frac{1}{3/5}=\)\(\frac{5}{3}\)
2. \(\text{sec}\,\theta=\)\(-\frac{13}{5}\)
3. \(\text{cot}\,\theta=\)\(\frac{24}{7}\)
4. \(\text{cot}\,\theta=\frac{1}{-2}=\)\(-\frac{1}{2}\)

Problem 3: Pythagorean forms

1. \(\sin^{2}\theta=1-0.6^{2}=1-0.36=\)\(0.64\)
2. \(\cos^{2}\theta=1-\frac{1}{4}=\)\(\frac{3}{4}\)
3. \(\text{sec}^{2}\theta=1+3^{2}=\)\(10\)
4. \(\text{csc}^{2}\theta=1+2^{2}=\)\(5\)

Problem 4: Cosine from sine, Quadrant I

\(\cos^{2}\theta=1-\frac{16}{25}=\frac{9}{25}\). Cosine is positive in Quadrant I, so \(\cos\theta=\frac{3}{5}\). Then \(\tan\theta=\frac{4/5}{3/5}=\)\(\frac{4}{3}\).

Problem 5: Cosine from sine, Quadrant II

\(\cos^{2}\theta=1-\frac{25}{169}=\frac{144}{169}\). Cosine is negative in Quadrant II, so \(\cos\theta=-\frac{12}{13}\). Then \(\tan\theta=\frac{5/13}{-12/13}=\)\(-\frac{5}{12}\).

Precalculus Trig Identities Worksheet Answer Key: problems 6–10

Problem 6: Sine from cosine, Quadrant III

\(\sin^{2}\theta=1-\frac{64}{289}=\frac{225}{289}\). Sine is negative in Quadrant III, so \(\sin\theta=-\frac{15}{17}\). Then \(\tan\theta=\frac{-15/17}{-8/17}=\)\(\frac{15}{8}\).

Problem 7: Sine and cosine from tangent

\(\text{sec}^{2}\theta=1+\frac{9}{16}=\frac{25}{16}\). Cosine is positive in Quadrant IV, so \(\text{sec}\,\theta=\frac{5}{4}\) and \(\cos\theta=\frac{4}{5}\). Then \(\sin\theta=\tan\theta\cos\theta=-\frac{3}{4}\cdot\frac{4}{5}=\)\(-\frac{3}{5}\).

Problem 8: All six functions

\(\cos^{2}\theta=1-\frac{576}{625}=\frac{49}{625}\), so \(\cos\theta=-\frac{7}{25}\). Then \(\tan\theta=\frac{24}{7}\), \(\text{csc}\,\theta=-\frac{25}{24}\), \(\text{sec}\,\theta=-\frac{25}{7}\), \(\text{cot}\,\theta=\frac{7}{24}\).

Problem 9: Simplify products

1. \(\sin\theta\cdot\frac{\cos\theta}{\sin\theta}=\)\(\cos\theta\)
2. \(\cos\theta\cdot\frac{\sin\theta}{\cos\theta}=\)\(\sin\theta\)
3. \(\frac{1}{\sin\theta}\cdot\frac{\sin\theta}{\cos\theta}=\frac{1}{\cos\theta}=\)\(\text{sec}\,\theta\)
4. \(1-\cos^{2}\theta=\)\(\sin^{2}\theta\)

Problem 10: Simplify with Pythagorean identities

1. \(1+\tan^{2}\theta=\)\(\text{sec}^{2}\theta\)
2. \(1-\sin^{2}\theta=\)\(\cos^{2}\theta\)
3. \(\frac{\tan^{2}\theta}{\tan\theta}=\)\(\tan\theta\)
4. \(\cos^{2}\theta\,\text{sec}^{2}\theta=\)\(1\)

Precalculus Trig Identities Worksheet Answer Key: problems 11–15

Problem 11: Identity or not?

1. The curves meet only at isolated points such as \(x=0\) and \(x=\frac{\pi}{2}\); an identity needs the graphs to match everywhere.
2. \(\sin\frac{\pi}{4}+\cos\frac{\pi}{4}=\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}=\sqrt{2}\approx 1.414\neq 1\), so the statement is false
3. \((\frac{\sqrt{3}}{2})^{2}+(\frac{1}{2})^{2}=\frac{3}{4}+\frac{1}{4}=\)\(1\)

Problem 12: Simplify rational expressions

1. \(\frac{1-\cos^{2}\theta}{1-\cos\theta}=\frac{(1-\cos\theta)(1+\cos\theta)}{1-\cos\theta}=\)\(1+\cos\theta\)
2. \(\frac{1-\sin^{2}\theta}{1-\sin\theta}=\)\(1+\sin\theta\)
3. \(\frac{(1+\sin\theta)+(1-\sin\theta)}{1-\sin^{2}\theta}=\frac{2}{\cos^{2}\theta}=\)\(2\,\text{sec}^{2}\theta\)

Problem 13: Verify an identity: sum of fractions

Combine the left side over \(\sin\theta(1+\cos\theta)\): \(\frac{\sin^{2}\theta+(1+\cos\theta)^{2}}{\sin\theta(1+\cos\theta)}\). The numerator is \(\sin^{2}\theta+1+2\cos\theta+\cos^{2}\theta=2+2\cos\theta=2(1+\cos\theta)\). Cancel \(1+\cos\theta\) to get \(\frac{2}{\sin\theta}=2\,\text{csc}\,\theta\). Verified.

Problem 14: Verify an identity: tangent plus cotangent

\(\frac{\sin\theta}{\cos\theta}+\frac{\cos\theta}{\sin\theta}=\frac{\sin^{2}\theta+\cos^{2}\theta}{\sin\theta\cos\theta}=\frac{1}{\sin\theta\cos\theta}=\text{sec}\,\theta\,\text{csc}\,\theta\). Verified.

Problem 15: Verify an identity: secant minus cosine

\(\text{sec}\,\theta-\cos\theta=\frac{1-\cos^{2}\theta}{\cos\theta}=\frac{\sin^{2}\theta}{\cos\theta}\). Dividing by \(\tan\theta=\frac{\sin\theta}{\cos\theta}\) gives \(\frac{\sin^{2}\theta}{\cos\theta}\cdot\frac{\cos\theta}{\sin\theta}=\sin\theta\). Verified.

Precalculus Trig Identities Worksheet Answer Key: problems 16–20

Problem 16: Exact values with the sum formula

1. \(\sin(45^{\circ}+30^{\circ})=\frac{\sqrt{2}}{2}\cdot\frac{\sqrt{3}}{2}+\frac{\sqrt{2}}{2}\cdot\frac{1}{2}=\)\(\frac{\sqrt{6}+\sqrt{2}}{4}\)
2. \(\cos(60^{\circ}+45^{\circ})=\frac{1}{2}\cdot\frac{\sqrt{2}}{2}-\frac{\sqrt{3}}{2}\cdot\frac{\sqrt{2}}{2}=\)\(\frac{\sqrt{2}-\sqrt{6}}{4}\)
3. \(\tan(45^{\circ}-30^{\circ})=\frac{1-\frac{\sqrt{3}}{3}}{1+\frac{\sqrt{3}}{3}}=\frac{3-\sqrt{3}}{3+\sqrt{3}}=\frac{(3-\sqrt{3})^{2}}{6}=\frac{12-6\sqrt{3}}{6}=\)\(2-\sqrt{3}\)

Problem 17: Exact values in radians

1. \(\frac{\pi}{12}=\frac{\pi}{4}-\frac{\pi}{6}\): \(\frac{\sqrt{2}}{2}\cdot\frac{\sqrt{3}}{2}-\frac{\sqrt{2}}{2}\cdot\frac{1}{2}=\)\(\frac{\sqrt{6}-\sqrt{2}}{4}\)
2. \(\frac{5\pi}{12}=\frac{\pi}{4}+\frac{\pi}{6}\): \(\frac{1+\frac{\sqrt{3}}{3}}{1-\frac{\sqrt{3}}{3}}=\frac{(3+\sqrt{3})^{2}}{6}=\frac{12+6\sqrt{3}}{6}=\)\(2+\sqrt{3}\)
3. \(\frac{7\pi}{12}=\frac{\pi}{3}+\frac{\pi}{4}\): \(\frac{1}{2}\cdot\frac{\sqrt{2}}{2}-\frac{\sqrt{3}}{2}\cdot\frac{\sqrt{2}}{2}=\)\(\frac{\sqrt{2}-\sqrt{6}}{4}\)

Problem 18: Sum formulas from two triangles

\(\cos A=\frac{4}{5}\) and \(\sin B=\frac{12}{13}\). Then \(\sin(A+B)=\frac{3}{5}\cdot(-\frac{5}{13})+\frac{4}{5}\cdot\frac{12}{13}=\frac{-15+48}{65}=\)\(\frac{33}{65}\) and \(\cos(A+B)=\frac{4}{5}\cdot(-\frac{5}{13})-\frac{3}{5}\cdot\frac{12}{13}=\frac{-20-36}{65}=\)\(-\frac{56}{65}\). Sine is positive and cosine negative, so A + B is in Quadrant II.

Problem 19: Recognize the formula

1. \(\sin(50^{\circ}-20^{\circ})=\sin 30^{\circ}=\)\(\frac{1}{2}\)
2. \(\cos(80^{\circ}-20^{\circ})=\cos 60^{\circ}=\)\(\frac{1}{2}\)
3. \(\tan(40^{\circ}+5^{\circ})=\tan 45^{\circ}=\)\(1\)
4. \(\sin(\frac{2\pi}{9}+\frac{\pi}{9})=\sin\frac{\pi}{3}=\)\(\frac{\sqrt{3}}{2}\)

Problem 20: Prove shift identities

1. \(\sin\theta\cos\frac{\pi}{2}+\cos\theta\sin\frac{\pi}{2}=\sin\theta\cdot 0+\cos\theta\cdot 1=\)\(\cos\theta\)
2. \(\cos\theta\cos\pi-\sin\theta\sin\pi=\cos\theta(-1)-\sin\theta(0)=\)\(-\cos\theta\)
3. \(\frac{\tan\theta+\tan\pi}{1-\tan\theta\tan\pi}=\frac{\tan\theta+0}{1-0}=\)\(\tan\theta\)

Precalculus Trig Identities Worksheet Answer Key: problems 21–25

Problem 21: Double-angle values

\(\cos\theta=\frac{12}{13}\) in Quadrant IV. Then \(\sin 2\theta=2(-\frac{5}{13})(\frac{12}{13})=\)\(-\frac{120}{169}\), \(\cos 2\theta=\frac{144}{169}-\frac{25}{169}=\)\(\frac{119}{169}\) and \(\tan 2\theta=\frac{-120/169}{119/169}=\)\(-\frac{120}{119}\).

Problem 22: Double-angle simplifications

1. \(\sin 30^{\circ}=\)\(\frac{1}{2}\)
2. \(\cos 45^{\circ}=\)\(\frac{\sqrt{2}}{2}\)
3. \(\cos\frac{\pi}{6}=\)\(\frac{\sqrt{3}}{2}\)
4. \(\tan 30^{\circ}=\)\(\frac{\sqrt{3}}{3}\)

Problem 23: Work backward from cos 2θ

\(2\cos^{2}\theta-1=\frac{7}{25}\) gives \(\cos^{2}\theta=\frac{16}{25}\), so \(\cos\theta=\frac{4}{5}\). \(1-2\sin^{2}\theta=\frac{7}{25}\) gives \(\sin^{2}\theta=\frac{9}{25}\), so \(\sin\theta=\frac{3}{5}\). Then \(\sin 2\theta=2\cdot\frac{3}{5}\cdot\frac{4}{5}=\frac{24}{25}\) and \(\tan 2\theta=\frac{24/25}{7/25}=\)\(\frac{24}{7}\).

Problem 24: Soccer kick range

1. \(\sin 2\theta=2(0.6)(0.8)=0.96\), so \(R=\frac{1600(0.96)}{32}=\)\(48\text{ ft}\)
2. \(\sin 90^{\circ}=1\), so \(R=\frac{1600}{32}=\)\(50\text{ ft}\)
3. \(2\text{ ft}\) farther.

Problem 25: Ferris wheel doubling

a) \(\sin\theta=\frac{30}{50}=\frac{3}{5}\) and \(\cos\theta=-\frac{4}{5}\) (left side). b) \(\cos 2\theta=\frac{16}{25}-\frac{9}{25}=\frac{7}{25}\) and \(\sin 2\theta=2\cdot\frac{3}{5}\cdot(-\frac{4}{5})=-\frac{24}{25}\). Multiply by 50: \((14,-48)\). c) \(60-48=\)\(12\text{ ft}\) above the ground.

Precalculus Trig Identities Worksheet Answer Key: problems 26–28

Problem 26: Rotating a point

1. \(\cos A=\frac{4}{5}\) and \(\sin A=\frac{3}{5}\)
2. \(x^{\prime}=5\cos(A+30^{\circ})=5(\frac{4}{5}\cdot\frac{\sqrt{3}}{2}-\frac{3}{5}\cdot\frac{1}{2})=\)\(2\sqrt{3}-\frac{3}{2}\approx 1.96\) and \(y^{\prime}=5(\frac{3}{5}\cdot\frac{\sqrt{3}}{2}+\frac{4}{5}\cdot\frac{1}{2})=\)\(\frac{3\sqrt{3}}{2}+2\approx 4.60\)
3. \(1.964^{2}+4.598^{2}\approx 3.857+21.143=25.000\), so the distance is 5

Problem 27: Camera tilt

1. \(\tan(A+B)=\frac{\frac{1}{2}+\frac{1}{3}}{1-\frac{1}{2}\cdot\frac{1}{3}}=\frac{5/6}{5/6}=\)\(1\)
2. Since \(\tan 45^{\circ}=1\) and \(A+B\) is acute, \(A+B=45^{\circ}\)
3. \(A\approx 26.57^{\circ}\) and \(B\approx 18.43^{\circ}\); \(26.57+18.43=45.00\). The check works.

Problem 28: Polarizing filter

1. Add 1 to both sides: \(1+\cos 2\theta=2\cos^{2}\theta\), then divide by 2.
2. \(I=80\cdot\frac{1+\frac{1}{2}}{2}=80\cdot\frac{3}{4}=\)\(60\text{ lumens}\)
3. \(I=80\cdot\frac{1+\cos 30^{\circ}}{2}=40(1+\frac{\sqrt{3}}{2})=\)\(40+20\sqrt{3}\approx 74.64\text{ lumens}\)
4. \(I=40(1+\cos 45^{\circ})=40(1+\frac{\sqrt{2}}{2})=\)\(40+20\sqrt{2}\approx 68.28\text{ lumens}\)

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