M
Math Coach Amy
Amy Ferguson Moncure

PreCalc 2 · Analytic Trigonometry

Half-Angle Formulas

Find an exact value of sin(θ/2), cos(θ/2), or tan(θ/2). The sign comes from the quadrant of the half angle.

Jump to section:

Key Ideas Easy Medium
What to Remember

Half-angle formulas

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

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

tan(θ/2) = sin θ1 + cos θ

Pick the sign

Look at the quadrant of θ/2, not the quadrant of θ.

If θ is between 0° and 180°, then θ/2 is in Quadrant I.

If θ is between 180° and 360°, then θ/2 is in Quadrant II.

In Quadrant II, sine is positive and cosine is negative.

A known half

22.5° is half of 45°.

15° is half of 30°.

Use the positive root. Both angles are in Quadrant I.

Tangent form

The tangent formula is sine over 1 + cosine.

That form already carries the sign. You do not choose ±.

Level 1

Easy

The half angle is in Quadrant I.

Example
22.5° is half of 45°. Find the exact value of cos 22.5°.

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

θ = 45°

22.5° is in Quadrant I, so use the positive square root.

cos 45° = square root of 22

1 + cos 45°

1 + square root of 22

22 + square root of 22

2 + square root of 22

Divide by 2.

2 + square root of 24

square root of 2 + square root of 24

square root of 2 + square root of 2square root of 4

square root of 4 = 2

square root of 2 + square root of 22

Answer: square root of 2 + square root of 22

1
22.5° is half of 45°. Find the exact value of sin 22.5°.
Show Answer

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

θ = 45°

22.5° is in Quadrant I, so use the positive square root.

cos 45° = square root of 22

1 − cos 45°

1 − square root of 22

22 − square root of 22

2 − square root of 22

Divide by 2.

2 − square root of 24

square root of 2 − square root of 24

square root of 2 − square root of 2square root of 4

square root of 4 = 2

square root of 2 − square root of 22

Answer: square root of 2 − square root of 22

2
22.5° is half of 45°. Find the exact value of tan 22.5°.
Show Answer

tan(θ/2) = 1 − cos θsin θ

θ = 45°

sin 45° = square root of 22

cos 45° = square root of 22

1 − cos 45°

2 − square root of 22

Divide by sin 45°.

2 − square root of 22square root of 22

2 − square root of 22 · 2square root of 2

2 − square root of 2square root of 2

2square root of 2 − square root of 2square root of 2

square root of 2 − 1

Answer: square root of 2 − 1

3
15° is half of 30°. Find the exact value of sin 15°.
Show Answer

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

θ = 30°

15° is in Quadrant I, so use the positive square root.

cos 30° = square root of 32

1 − cos 30°

1 − square root of 32

22 − square root of 32

2 − square root of 32

Divide by 2.

2 − square root of 34

square root of 2 − square root of 34

square root of 2 − square root of 3square root of 4

square root of 4 = 2

square root of 2 − square root of 32

Answer: square root of 2 − square root of 32

4
15° is half of 30°. Find the exact value of cos 15°.
Show Answer

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

θ = 30°

15° is in Quadrant I, so use the positive square root.

cos 30° = square root of 32

1 + cos 30°

1 + square root of 32

22 + square root of 32

2 + square root of 32

Divide by 2.

2 + square root of 34

square root of 2 + square root of 34

square root of 2 + square root of 3square root of 4

square root of 4 = 2

square root of 2 + square root of 32

Answer: square root of 2 + square root of 32

5
15° is half of 30°. Find the exact value of tan 15°.
Show Answer

tan(θ/2) = 1 − cos θsin θ

θ = 30°

sin 30° = 12

cos 30° = square root of 32

1 − cos 30°

2 − square root of 32

Divide by sin 30°.

2 − square root of 3212

2 − square root of 32 · 2

2 − square root of 3

Answer: 2 − square root of 3

6
cos θ = 35 and θ is in Quadrant I. 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 = 35

1 − cos θ

1 − 35

55 − 35

25

252

25 · 12

15

square root of 15

square root of 55

Answer: square root of 55

7
cos θ = 35 and θ is in Quadrant I. 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 = 35

1 + cos θ

1 + 35

55 + 35

85

852

85 · 12

45

square root of 45

2square root of 55

Answer: 2square root of 55

8
cos θ = 725 and θ is in Quadrant I. 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 = 725

1 − cos θ

1 − 725

2525 − 725

1825

18252

1825 · 12

925

square root of 925

35

Answer: 35

9
cos θ = 725 and θ is in Quadrant I. 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 = 725

1 + cos θ

1 + 725

2525 + 725

3225

32252

3225 · 12

1625

square root of 1625

45

Answer: 45

10
cos θ = 35 and θ is in Quadrant I. 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 = 35

cos² θ = 925

1 − 925

2525 − 925

1625

sin² θ = 1625

θ is in Quadrant I, so sine is positive.

sin θ = 45

1 + cos θ

1 + 35

55 + 35

85

4585

45 · 58

12

Answer: 12

Level 2

Medium

Choose the sign from the quadrant of θ/2.

Example
cos θ = −35 and θ is in Quadrant III. Find cos(θ/2).

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 = −35

1 + cos θ

1 + (−35)

1 − 35

25

252

25 · 12

15

square root of 15

−square root of 55

Answer: −square root of 55

1
cos θ = −35 and θ is in Quadrant II. Find sin(θ/2).
Show Answer

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

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

Use the positive root.

cos θ = −35

1 − cos θ

1 − (−35)

1 + 35

55 + 35

85

852

85 · 12

45

square root of 45

2square root of 55

Answer: 2square root of 55

2
cos θ = −35 and θ is in Quadrant III. Find sin(θ/2).
Show Answer

sin(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 positive root.

cos θ = −35

1 − cos θ

1 − (−35)

1 + 35

55 + 35

85

852

85 · 12

45

square root of 45

2square root of 55

Answer: 2square root of 55

3
cos θ = 513 and θ is in Quadrant IV. Find cos(θ/2).
Show Answer

cos(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 negative root for cosine.

cos θ = 513

1 + cos θ

1 + 513

1313 + 513

1813

18132

1813 · 12

913

square root of 913

−3square root of 1313

Answer: −3square root of 1313

4
cos θ = 513 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 = 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

5
cos θ = −725 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 = −725

1 + cos θ

1 + (−725)

1 − 725

1825

18252

1825 · 12

925

square root of 925

−35

Answer: −35

6
cos θ = −725 and θ is in Quadrant III. Find sin(θ/2).
Show Answer

sin(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 positive root.

cos θ = −725

1 − cos θ

1 − (−725)

1 + 725

2525 + 725

3225

32252

3225 · 12

1625

square root of 1625

45

Answer: 45

7
cos θ = −45 and θ is in Quadrant III. Find sin(θ/2).
Show Answer

sin(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 positive root.

cos θ = −45

1 − cos θ

1 − (−45)

1 + 45

55 + 45

95

952

95 · 12

910

square root of 910

3square root of 1010

Answer: 3square root of 1010

8
cos θ = −45 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 = −45

cos² θ = 1625

1 − 1625

2525 − 1625

925

sin² θ = 925

θ is in Quadrant II, so sine is positive.

sin θ = 35

1 + cos θ

1 + (−45)

1 − 45

15

3515

35 · 5

3

Answer: 3

9
cos θ = 513 and θ is in Quadrant IV. Find tan(θ/2).
Show Answer

tan(θ/2) = sin θ1 + cos θ

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

sin θ = −1213

1 + cos θ

1 + 513

1313 + 513

1813

−12131813

−1213 · 1318

−23

Answer: −23

10
cos θ = −817 and θ is in Quadrant II. Find cos(θ/2).
Show Answer

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

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

Use the positive root.

cos θ = −817

1 + cos θ

1 + (−817)

1 − 817

917

9172

917 · 12

934

square root of 934

3square root of 3434

Answer: 3square root of 3434

Need a little more help with this skill?

Book a Session with Amy