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Amy Ferguson Moncure

PreCalc 2 · Analytic Trigonometry

Mixed Practice: Angle Formulas

Choose a sum, a difference, a double-angle, a half-angle, or a power-reducing formula. Then find the exact value.

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Key Ideas Easy Medium
What to Remember

Which formula

A sum or difference of 30°, 45°, or 60° uses a sum or difference formula.

2 sin θ cos θ, or a square minus a square, is a double angle.

An angle that is half of one you know uses a half-angle formula.

A square of sine, cosine, or tangent uses a power-reducing formula.

Sum and difference

sin(A + B) = sin A cos B + cos A sin B

cos(A − B) = cos A cos B + sin A sin B

tan(A + B) = tan A + tan B1 − tan A · tan B

Double angle

sin 2θ = 2 sin θ cos θ

cos 2θ = 1 − 2 sin² θ

tan 2θ = 2 tan θ1 − tan² θ

Half angle

sin(θ/2) = ± square root of 1 − cos θ2

tan(θ/2) = sin θ1 + cos θ

Read the sign from the quadrant of θ/2.

Power reducing

sin² θ = 1 − cos 2θ2

cos² θ = 1 + cos 2θ2

tan² θ = 1 − cos 2θ1 + cos 2θ

A square needs no ±.

Level 1

Easy

The formula to use is named in the problem.

Example
Use a sum formula to find the exact value of tan 165°.

tan(A + B) = tan A + tan B1 − tan A · tan B

A = 120°

B = 45°

tan 120° = −square root of 3

tan 45° = 1

tan 120° + tan 45°

1 − square root of 3

tan 120° · tan 45°

−square root of 3

1 − tan A · tan B

1 + square root of 3

square root of 3 − 2

Answer: square root of 3 − 2

1
Use a sum formula to find the exact value of sin 150°.
Show Answer

sin(A + B) = sin A cos B + cos A sin B

A = 120°

B = 30°

sin 120° · cos 30° + cos 120° · sin 30°

sin 120° = square root of 32

cos 30° = square root of 32

cos 120° = −12

sin 30° = 12

sin 120° · cos 30°

34

cos 120° · sin 30°

−14

12

Answer: 12

2
Use a difference formula to find the exact value of cos(120° − 45°).
Show Answer

cos(A − B) = cos A cos B + sin A sin B

A = 120°

B = 45°

cos 120° · cos 45° + sin 120° · sin 45°

cos 120° = −12

cos 45° = square root of 22

sin 120° = square root of 32

sin 45° = square root of 22

cos 120° · cos 45°

−square root of 24

sin 120° · sin 45°

square root of 64

square root of 6 − square root of 24

Answer: square root of 6 − square root of 24

3
Use a double-angle formula to find the exact value of 2 sin 15° cos 15°.
Show Answer

sin 2θ = 2 sin θ cos θ

2 sin 15° cos 15° is sin 30°.

sin 30° = 12

Answer: 12

4
Use a double-angle formula to find the exact value of 1 − 2 sin² 15°.
Show Answer

cos 2θ = 1 − 2 sin² θ

1 − 2 sin² 15° is cos 30°.

cos 30° = square root of 32

Answer: square root of 32

5
Use a power-reducing formula to find the exact value of sin² 105°.
Show Answer

sin² θ = 1 − cos 2θ2

θ = 105°

2θ = 210°

cos 210° = −square root of 32

1 − cos 210°

1 − (−square root of 32)

1 + square root of 32

22 + square root of 32

2 + square root of 32

Divide by 2.

2 + square root of 322

2 + square root of 32 · 12

2 + square root of 34

Answer: 2 + square root of 34

6
cos θ = 513 and θ is in Quadrant I. Use a half-angle formula to find sin(θ/2).
Show Answer

sin(θ/2) = ± square root of 1 − cos θ2

θ is in Quadrant I, so θ/2 is in Quadrant I. Sine and cosine are both positive there.

Use the positive root.

cos θ = 513

1 − cos θ

1 − 513

1313 − 513

813

8132

813 · 12

413

square root of 413

2square root of 1313

Answer: 2square root of 1313

7
cos θ = 513 and θ is in Quadrant I. Use a half-angle formula to find tan(θ/2).
Show Answer

tan(θ/2) = sin θ1 + cos θ

θ is in Quadrant I, so θ/2 is in Quadrant I. Sine and cosine are both positive there.

sin² θ + cos² θ = 1

sin² θ = 1 − cos² θ

cos θ = 513

cos² θ = 25169

1 − 25169

169169 − 25169

144169

sin² θ = 144169

θ is in Quadrant I, so sine is positive.

sin θ = 1213

1 + cos θ

1 + 513

1313 + 513

1813

12131813

1213 · 1318

23

Answer: 23

8
sin θ = 817 and cos θ = 1517. Use a double-angle formula to find sin 2θ.
Show Answer

sin 2θ = 2 sin θ cos θ

sin θ = 817

cos θ = 1517

2 · 817 · 1517

240289

Answer: 240289

9
tan θ = 34. Use a double-angle formula to find tan 2θ.
Show Answer

tan 2θ = 2 tan θ1 − tan² θ

tan θ = 34

2 tan θ

32

tan² θ = 916

1 − tan² θ

1 − 916

1616 − 916

716

32716

32 · 167

247

Answer: 247

10
cos 2θ = 1517. Use a power-reducing formula to find cos² θ.
Show Answer

cos² θ = 1 + cos 2θ2

cos 2θ = 1517

1 + cos 2θ

1 + 1517

1717 + 1517

3217

Divide by 2.

32172

3217 · 12

3234

1617

Answer: 1617

11
cos θ = 1213 and θ is in Quadrant I. Use a half-angle formula to find cos(θ/2).
Show Answer

cos(θ/2) = ± square root of 1 + cos θ2

θ is in Quadrant I, so θ/2 is in Quadrant I. Sine and cosine are both positive there.

Use the positive root.

cos θ = 1213

1 + cos θ

1 + 1213

1313 + 1213

2513

25132

2513 · 12

2526

square root of 2526

5square root of 2626

Answer: 5square root of 2626

12
Use a sum formula to find the exact value of cos 150°.
Show Answer

cos(A + B) = cos A cos B − sin A sin B

A = 120°

B = 30°

cos 120° · cos 30° − sin 120° · sin 30°

cos 120° = −12

cos 30° = square root of 32

sin 120° = square root of 32

sin 30° = 12

cos 120° · cos 30°

−square root of 34

sin 120° · sin 30°

square root of 34

−square root of 32

Answer: −square root of 32

Level 2

Medium

Choose the formula. A square needs no sign. Watch the quadrant of a half angle.

Example
Find the exact value of cos(A + B). sin A = 725 and A is in Quadrant I. cos B = 817 and B is in Quadrant I.

cos(A + B) = cos A cos B − sin A sin B

sin² A + cos² A = 1

cos² A = 1 − sin² A

sin A = 725

sin² A = 49625

1 − 49625

625625 − 49625

576625

cos² A = 576625

A is in Quadrant I, so cosine is positive.

cos A = 2425

sin² B + cos² B = 1

sin² B = 1 − cos² B

cos B = 817

cos² B = 64289

1 − 64289

289289 − 64289

225289

sin² B = 225289

B is in Quadrant I, so sine is positive.

sin B = 1517

cos A · cos B − sin A · sin B

cos A · cos B

2425 · 817

192425

sin A · sin B

725 · 1517

105425

2185

192425 − 2185

192425 − 105425

87425

Answer: 87425

1
Find the exact value of sin(A + B). sin A = 2029 and A is in Quadrant I. cos B = 817 and B is in Quadrant I.
Show Answer

sin(A + B) = sin A cos B + cos A sin B

sin² A + cos² A = 1

cos² A = 1 − sin² A

sin A = 2029

sin² A = 400841

1 − 400841

841841 − 400841

441841

cos² A = 441841

A is in Quadrant I, so cosine is positive.

cos A = 2129

sin² B + cos² B = 1

sin² B = 1 − cos² B

cos B = 817

cos² B = 64289

1 − 64289

289289 − 64289

225289

sin² B = 225289

B is in Quadrant I, so sine is positive.

sin B = 1517

sin A · cos B + cos A · sin B

sin A · cos B

2029 · 817

160493

cos A · sin B

2129 · 1517

315493

160493 + 315493

475493

Answer: 475493

2
Find the exact value of tan(A − B). tan A = 52. tan B = 13.
Show Answer

tan(A − B) = tan A − tan B1 + tan A · tan B

tan A = 52

tan B = 13

tan A − tan B

52 − 13

156 − 26

136

tan A · tan B

52 · 13

56

1 + tan A · tan B

1 + 56

66 + 56

116

136116

136 · 611

1311

Answer: 1311

3
sin θ = 2029 and θ is in Quadrant II. Find sin 2θ.
Show Answer

sin 2θ = 2 sin θ cos θ

sin² θ + cos² θ = 1

cos² θ = 1 − sin² θ

sin θ = 2029

sin² θ = 400841

1 − 400841

841841 − 400841

441841

cos² θ = 441841

θ is in Quadrant II, so cosine is negative.

cos θ = −2129

sin θ = 2029

cos θ = −2129

2 · 2029 · (−2129)

−840841

Answer: −840841

4
Find the exact value of cos² 112.5°.
Show Answer

cos² θ = 1 + cos 2θ2

θ = 112.5°

2θ = 225°

cos 225° = −square root of 22

1 + cos 225°

1 + (−square root of 22)

1 − square root of 22

22 − square root of 22

2 − square root of 22

Divide by 2.

2 − square root of 222

2 − square root of 22 · 12

2 − square root of 24

Answer: 2 − square root of 24

5
cos θ = −1213 and θ is in Quadrant III. Find cos(θ/2).
Show Answer

cos(θ/2) = ± square root of 1 + cos θ2

θ is in Quadrant III, so θ/2 is in Quadrant II. Sine is positive there. Cosine is negative there.

Use the negative root for cosine.

cos θ = −1213

1 + cos θ

1 + (−1213)

1 − 1213

113

1132

113 · 12

126

square root of 126

−square root of 2626

Answer: −square root of 2626

6
Find the exact value of sin(120° − 45°).
Show Answer

sin(A − B) = sin A cos B − cos A sin B

A = 120°

B = 45°

sin 120° · cos 45° − cos 120° · sin 45°

sin 120° = square root of 32

cos 45° = square root of 22

cos 120° = −12

sin 45° = square root of 22

sin 120° · cos 45°

square root of 64

cos 120° · sin 45°

−square root of 24

square root of 6 + square root of 24

Answer: square root of 6 + square root of 24

7
cos θ = 817 and θ is in Quadrant IV. Find sin(θ/2).
Show Answer

sin(θ/2) = ± square root of 1 − cos θ2

θ is in Quadrant IV, so θ/2 is in Quadrant II. Sine is positive there. Cosine is negative there.

Use the positive root.

cos θ = 817

1 − cos θ

1 − 817

1717 − 817

917

9172

917 · 12

934

square root of 934

3square root of 3434

Answer: 3square root of 3434

8
cos θ = −513 and θ is in Quadrant II. Find cos 2θ.
Show Answer

cos 2θ = cos² θ − sin² θ

sin² θ + cos² θ = 1

sin² θ = 1 − cos² θ

cos θ = −513

cos² θ = 25169

1 − 25169

169169 − 25169

144169

sin² θ = 144169

θ is in Quadrant II, so sine is positive.

sin θ = 1213

cos θ = −513

cos² θ = 25169

sin θ = 1213

sin² θ = 144169

25169 − 144169

−119169

Answer: −119169

9
cos 2θ = −13. Find the exact value of tan² θ.
Show Answer

tan² θ = 1 − cos 2θ1 + cos 2θ

cos 2θ = −13

1 − cos 2θ

1 − (−13)

1 + 13

33 + 13

43

1 + cos 2θ

1 + (−13)

1 − 13

33 − 13

23

4323

43 · 32

2

Answer: 2

10
Find the exact value of sin(A − B). sin A = 1213 and A is in Quadrant II. cos B = −35 and B is in Quadrant III.
Show Answer

sin(A − B) = sin A cos B − cos A sin B

sin² A + cos² A = 1

cos² A = 1 − sin² A

sin A = 1213

sin² A = 144169

1 − 144169

169169 − 144169

25169

cos² A = 25169

A is in Quadrant II, so cosine is negative.

cos A = −513

sin² B + cos² B = 1

sin² B = 1 − cos² B

cos B = −35

cos² B = 925

1 − 925

2525 − 925

1625

sin² B = 1625

B is in Quadrant III, so sine is negative.

sin B = −45

sin A · cos B − cos A · sin B

sin A · cos B

1213 · (−35)

−3665

cos A · sin B

(−513) · (−45)

2065

413

−3665 − 413

−3665 − 2065

−5665

Answer: −5665

11
tan θ = −815. Find tan 2θ.
Show Answer

tan 2θ = 2 tan θ1 − tan² θ

tan θ = −815

2 tan θ

−1615

tan² θ = 64225

1 − tan² θ

1 − 64225

225225 − 64225

161225

−1615161225

−1615 · 225161

−240161

Answer: −240161

12
cos θ = −725 and θ is in Quadrant II. Find tan(θ/2).
Show Answer

tan(θ/2) = sin θ1 + cos θ

θ is in Quadrant II, so θ/2 is in Quadrant I. Sine and cosine are both positive there.

sin² θ + cos² θ = 1

sin² θ = 1 − cos² θ

cos θ = −725

cos² θ = 49625

1 − 49625

625625 − 49625

576625

sin² θ = 576625

θ is in Quadrant II, so sine is positive.

sin θ = 2425

1 + cos θ

1 + (−725)

1 − 725

1825

24251825

2425 · 2518

43

Answer: 43

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