Grade by Grade
D. Trigonometric Functions · Trigonometric Ratios, Identities and Laws

Solving for Angles: Two Angles with the Same Ratio

D1.3 D1.2About 70 minutes
Learning goals
  • I can explain why a calculator gives only one angle for sin⁻¹, cos⁻¹ and tan⁻¹.
  • I can find both angles from 0° to 360° that have a given sine, cosine or tangent.
  • I can solve simple trigonometric equations such as 2 cos θ + 1 = 0 for 0° ≤ θ ≤ 360°.
  • I can recognize equations with one solution or no solution.

Words to know

trigonometric equation
An equation that contains a trigonometric ratio of an unknown angle, such as sin θ = 0.4.
principal value
The single angle a calculator returns for sin⁻¹, cos⁻¹ or tan⁻¹. sin⁻¹ and tan⁻¹ give angles from −90° to 90°; cos⁻¹ gives angles from 0° to 180°.
interval
The set of allowed angles, written like 0° ≤ θ ≤ 360°.

One ratio, two angles

In Grade 10 you used sin⁻¹ to find an angle in a right triangle, and there was only one answer. Now angles can go all the way to 360°, and you saw in lesson 3 that sin 30° = sin 150° = 1/2. So if all you know is sin θ = 1/2, θ could be 30° or 150°. In this lesson you will learn to find both angles every time.

Why two?

On the unit circle, sin θ is the y-coordinate. The points with y = 1/2 are where the horizontal line y = 1/2 crosses the circle, and a horizontal line crosses a circle in two places (unless it just touches the top or bottom).

In the same way, cos θ is the x-coordinate, so the solutions of cos θ = k are where a vertical line x = k crosses the circle.

y = 1/230°150°sin θ = 1/2: θ = 30° or 150°x = −√2/2135°225°cos θ = −√2/2: θ = 135° or 225°
Left: the line y = 1/2 meets the unit circle at 30° and 150°. Right: the line x = −√2/2 meets it at 135° and 225°.

Steps for solving sin θ = k, cos θ = k or tan θ = k (0° ≤ θ ≤ 360°)

  1. Related acute angle: β = sin⁻¹|k| (or cos⁻¹|k|, tan⁻¹|k|). Use the size of k, without its sign.
  2. Quadrants: use the sign of k and CAST to pick the two quadrants.
  3. Angles: quadrant I: β; II: 180° − β; III: 180° + β; IV: 360° − β.
  4. Check each answer with a calculator.
Worked example

Example 1: An exact value

Solve sin θ = 1/2 for 0° ≤ θ ≤ 360°.

  1. β: sin β = 1/2, so β = 30°.
  2. Sine is positive in quadrants I and II.
  3. θ = 30° or θ = 180° − 30° = 150°.

θ = 30° or 150°

Worked example

Example 2: A negative exact value

Solve cos θ = −√2/2 for 0° ≤ θ ≤ 360°.

  1. β: cos β = √2/2, so β = 45°.
  2. Cosine is negative in quadrants II and III.
  3. θ = 180° − 45° = 135° or θ = 180° + 45° = 225°.

θ = 135° or 225°

Worked example

Example 3: Using a calculator

Solve tan θ = −1.5 for 0° ≤ θ ≤ 360°, to the nearest tenth of a degree.

  1. β = tan⁻¹(1.5) ≈ 56.3°. (If you type tan⁻¹(−1.5), the calculator gives −56.3°, which is not in our interval.)
  2. Tangent is negative in quadrants II and IV.
  3. θ ≈ 180° − 56.3° = 123.7° or θ ≈ 360° − 56.3° = 303.7°.
  4. Check: tan 123.7° ≈ −1.50 ✓ and tan 303.7° ≈ −1.50 ✓

θ ≈ 123.7° or 303.7°

xyθ ≈ 123.7°θ ≈ 303.7°ββtan θ = −1.5β = tan⁻¹(1.5) ≈ 56.3°Tangent is negative inquadrants II and IV:180° − 56.3° = 123.7°360° − 56.3° = 303.7°
The two solutions of tan θ = −1.5 lie on one straight line through the origin, so they are exactly 180° apart. Both make the same angle β with the x-axis.
Worked example

Example 4: A negative sine

Solve sin θ = −0.4 for 0° ≤ θ ≤ 360°, to the nearest tenth of a degree.

  1. β = sin⁻¹(0.4) ≈ 23.6°.
  2. Sine is negative in quadrants III and IV.
  3. θ ≈ 180° + 23.6° = 203.6° or θ ≈ 360° − 23.6° = 336.4°.

θ ≈ 203.6° or 336.4°

Worked example

Example 5: Isolate the ratio first

Solve 2 cos θ + 1 = 0 for 0° ≤ θ ≤ 360°.

  1. Treat cos θ like a variable: 2 cos θ = −1, so cos θ = −1/2.
  2. β = 60° (cos 60° = 1/2).
  3. Cosine is negative in quadrants II and III: θ = 120° or 240°.

θ = 120° or 240°

Special cases

  • On an axis: sin θ = 1 has only one solution, θ = 90° (the line y = 1 just touches the top of the circle). sin θ = 0 has three solutions in 0° ≤ θ ≤ 360°: 0°, 180° and 360°.
  • Impossible values: sin θ and cos θ are always between −1 and 1, because |y| ≤ r and |x| ≤ r. So sin θ = 1.2 has no solution. (tan θ can be any real number.)
Just for fun

The single angle your calculator returns for sin⁻¹ is called the principal value. So when the calculator gives only one answer, it isn't being lazy. It's just following the principal's rules.

Hands-on activity

Find both angles with a ruler

You need: the unit circle you drew in lesson 3 (radius 10 cm), a ruler, a protractor and a pencil.

  1. Lay the ruler horizontally across the circle, 4 cm above the x-axis. This is the line y = 0.4.
  2. Mark the two points where the ruler crosses the circle.
  3. Measure each angle in standard position with the protractor.
  4. Now place the ruler 4 cm below the x-axis (y = −0.4) and repeat.
  5. Finally, place the ruler vertically 6 cm to the left of the y-axis (x = −0.6) and measure both angles.

What to look for: for y = 0.4 you should measure about 24° and 156°; for y = −0.4, about 204° and 336°; for x = −0.6, about 127° and 233° (since cos⁻¹(0.6) ≈ 53.1°).

Your turn

1

Solve cos θ = 1/2 for 0° ≤ θ ≤ 360°.

2

Solve sin θ = −√3/2 for 0° ≤ θ ≤ 360°.

3

Solve tan θ = √3 for 0° ≤ θ ≤ 360°.

4

Solve cos θ = −0.8 to the nearest tenth of a degree.

5

Solve sin θ = 0.75 to the nearest tenth of a degree.

6

Solve tan θ = −0.25 to the nearest tenth of a degree.

7

Solve sin θ = −1 for 0° ≤ θ ≤ 360°.

More practice: warm-up

1

Solve sin θ = √2/2 for 0° ≤ θ ≤ 360°.

2

Solve cos θ = 0 for 0° ≤ θ ≤ 360°.

3

Solve tan θ = 1 for 0° ≤ θ ≤ 360°.

4

In which quadrants is cos θ negative?

5

Solve cos θ = 1 for 0° ≤ θ ≤ 360°.

More practice: on your own

1

Solve 2 sin θ − 1 = 0 for 0° ≤ θ ≤ 360°.

2

Solve 3 tan θ + 3 = 0 for 0° ≤ θ ≤ 360°.

3

Solve 5 cos θ = −2 to the nearest tenth of a degree.

4

Solve √2 sin θ + 1 = 0 for 0° ≤ θ ≤ 360°.

5

Mia solved sin θ = 0.6 and wrote only θ ≈ 36.9°. What did she miss?

Challenge

1

Solve sin²θ = 1/4 for 0° ≤ θ ≤ 360°.

2

Solve 2cos²θ − cos θ − 1 = 0 for 0° ≤ θ ≤ 360°. (Hint: factor like a quadratic.)

3

Explain why sin θ = 1.2 has no solution, but tan θ = 1.2 does.

Check your understanding

Quiz

1.

What are the solutions of sin θ = 1/2 for 0° ≤ θ ≤ 360°?

2.

In which quadrants is cos θ negative?

3.

What are the solutions of tan θ = √3 for 0° ≤ θ ≤ 360°?

4.

A calculator gives sin⁻¹(0.4) ≈ 23.6°. What is the other solution of sin θ = 0.4 between 0° and 360°?

5.

How many solutions does sin θ = −1 have for 0° ≤ θ ≤ 360°?

6.

What are the solutions of 2 cos θ + 1 = 0 for 0° ≤ θ ≤ 360°?

For parents and teachers

Big idea (D1.3): a sine, cosine or tangent value usually belongs to two angles between 0° and 360°. Students find the related acute angle with the calculator (using the positive value), then use CAST to place the two answers.

How to help: After your teen solves an equation, have them check each answer by typing it back into the calculator, for example sin 150° should give 0.5. Two checks, two answers.

Common difficulties:

  • Stopping after the calculator's single answer.
  • Typing the negative value into the inverse key (tan⁻¹(−1.5) = −56.3°) and not knowing what to do with a negative angle. Use the positive value to find β.
  • Using 180° − β for quadrant III (it should be 180° + β).
  • Not noticing that sin θ = 1.2 or cos θ = −3 is impossible.