C. Trigonometric Functions · Sine FunctionsAngles of Rotation and the Sine Ratio from 0° to 360°
C2.3About 70 minutes
Learning goals- I can draw an angle in standard position and name the quadrant of its terminal arm.
- I can find sin θ from a point on the terminal arm using sin θ = y ÷ r.
- I can explain why sin θ is positive in quadrants I and II and negative in III and IV.
- I can use the unit circle and related acute angles to find angles with a given sine.
Words to know
standard position
An angle with its vertex at the origin and its initial arm on the positive x-axis, opening counterclockwise.
initial arm / terminal arm
The starting ray (on the positive x-axis) and the ending ray of an angle in standard position.
quadrant
One of the four regions of the coordinate plane, numbered I to IV counterclockwise from the top right.
unit circle
A circle with radius 1 centred at the origin. For a point on it, sin θ = y.
related acute angle
The acute angle between the terminal arm and the x-axis (also called the reference angle).
A seat on a Ferris wheel
Picture a seat on a Ferris wheel at a fall fair. As the wheel turns, the seat's height above the centre of the wheel changes: positive when the seat is above the centre, zero level with it, negative below it. In Grade 10 the sine ratio worked only for angles less than 90° inside a right triangle. To describe a whole turn of the wheel, we need a sine for every angle from 0° to 360°. That is today's goal.
Remember from Grade 10
In a right triangle, sin θ = opposite ÷ hypotenuse. For example, if the opposite side is 3 and the hypotenuse is 5, then sin θ = 0.6, and θ = sin⁻¹(0.6) ≈ 36.9°. Make sure your calculator is in degree mode (sin 30° should give 0.5).
Angles in standard position
An angle is in standard position when:
- its vertex is at the origin (0, 0),
- its initial arm lies along the positive x-axis, and
- it opens counterclockwise to its terminal arm.
The axes split the plane into four quadrants, numbered I, II, III, IV counterclockwise from the top right. An angle between 0° and 90° ends in quadrant I, between 90° and 180° in quadrant II, between 180° and 270° in quadrant III, and between 270° and 360° in quadrant IV.
The sine of any angle
Pick any point P(x, y) on the terminal arm (not the origin). Its distance from the origin is
r = √(x² + y²), and r is always positive.
Then we define
sin θ = y ÷ r
For an angle in quadrant I, y is the opposite side and r is the hypotenuse of a right triangle, so this matches the Grade 10 definition. But now it works for every angle. Since r > 0, the sign of sin θ is the sign of y: positive above the x-axis (quadrants I and II), negative below it (quadrants III and IV).
Worked exampleExample 1: Sine from a point
The point P(−3, 4) is on the terminal arm of angle θ in standard position, with 0° ≤ θ ≤ 360°. Find sin θ and the measure of θ.
- Find r: r = √((−3)² + 4²) = √(9 + 16) = √25 = 5.
- sin θ = y ÷ r = 4 ÷ 5 = 0.8.
- P is in quadrant II (x negative, y positive), so θ is between 90° and 180°.
- The calculator's sin⁻¹(0.8) gives 53.1°, which is in quadrant I. The terminal arm in quadrant II makes the same 53.1° angle with the x-axis, but measured back from 180°: θ = 180° − 53.1° = 126.9°.
- Check: sin 126.9° ≈ 0.800 ✓.
sin θ = 0.8 and θ ≈ 126.9°.
The unit circle
If we choose P on a circle of radius r = 1 (a unit circle), then sin θ = y ÷ 1 = y. So:
sin θ is the height of the point above (or below) the x-axis as it travels around a circle of radius 1.
This is the key idea of the unit. As θ goes from 0° to 360°, the point goes once around the circle, and its height goes up to 1, back down to 0, down to −1 and back to 0.
Symmetry of the circle
The circle is symmetric, so the sines of different angles are related. If θ is an acute angle:
- sin(180° − θ) = sin θ (mirror image across the y-axis: same height). Example: sin 150° = sin 30° = 0.5.
- sin(180° + θ) = −sin θ (half a turn further: same distance, below the axis). Example: sin 210° = −0.5.
- sin(360° − θ) = −sin θ (mirror image across the x-axis). Example: sin 330° = −0.5.
The acute angle that the terminal arm makes with the x-axis is called the related acute angle (or reference angle). This idea is how we found 126.9° in Example 1.
| θ | quadrant | related acute angle | sin θ (calculator, 3 decimals) |
|---|
| 40° | I | 40° | 0.643 |
| 140° | II | 40° | 0.643 |
| 220° | III | 40° | −0.643 |
| 320° | IV | 40° | −0.643 |
Worked exampleExample 2: Heights on a Ferris wheel
A Ferris wheel has radius 10 m and its centre is 12 m above the ground. A seat starts level with the centre on the right (θ = 0°) and the wheel turns counterclockwise. Its height above the ground is h = 12 + 10 sin θ. Find the height after the wheel turns (a) 150°, (b) 240°, (c) 360°.
- The seat is on a circle of radius 10, so its height above the centre is y = 10 sin θ (from sin θ = y ÷ r with r = 10). Add 12 m for the centre.
- (a) sin 150° = 0.5, so h = 12 + 10(0.5) = 17 m.
- (b) sin 240° ≈ −0.866, so h ≈ 12 + 10(−0.866) = 12 − 8.66 ≈ 3.3 m (below the centre).
- (c) sin 360° = 0, so h = 12 + 0 = 12 m: the seat is back where it started.
About 17 m, 3.3 m and 12 m.
Just for funA circle walks into a geometry class and asks, "Is there a point to all this?"
The teacher says, "There are infinitely many, and every one of them has a sine."
Hands-on activityMake a unit circle with a paper plate
You need: a paper plate (or a circle traced from a pot lid), a ruler, a protractor, a pencil and a calculator.
- Mark the centre. Draw an x-axis and a y-axis through it.
- Measure the radius, r, in centimetres.
- Use the protractor to mark points on the edge at 30°, 60°, 120°, 150°, 210°, 240°, 300° and 330°, measured counterclockwise from the positive x-axis.
- For each point, measure its height above (+) or below (−) the x-axis, y, in cm.
- Calculate y ÷ r and compare it with sin θ on your calculator.
What to look for: y ÷ r should match the calculator to about 0.05 (measurement error). The values for 30° and 150° should match each other, and so should 60° and 120°. The values for 210°, 240°, 300° and 330° should be negative.
Your turn
1In which quadrant does the terminal arm of each angle lie? (a) 100° (b) 250° (c) 345°
2P(5, −12) is on the terminal arm of θ. Find r and sin θ.
r = √(25 + 144) = √169 = 13. sin θ = −12 ÷ 13 ≈ −0.923.
3Without a calculator, is sin 200° positive or negative? Explain.
Negative. 200° is in quadrant III, where y < 0, so sin θ = y ÷ r < 0.
4Use your calculator: sin 120°, sin 180°, sin 270°.
sin 120° ≈ 0.866, sin 180° = 0, sin 270° = −1.
5sin 35° ≈ 0.574. Write another angle between 0° and 360° with the same sine, and two angles whose sine is −0.574.
Same sine: 180° − 35° = 145°. Sine −0.574: 180° + 35° = 215° and 360° − 35° = 325°.
6Using the Ferris wheel in Example 2, find the height of the seat after a turn of 90° and after 270°.
90°: h = 12 + 10(1) = 22 m (top). 270°: h = 12 + 10(−1) = 2 m (bottom).
More practice: warm-up
1Where is the vertex of an angle in standard position?
2Which way do angles in standard position open?
Counterclockwise from the positive x-axis.
3On a unit circle, what is sin θ equal to?
The y-coordinate of the point.
5In which quadrants is sin θ negative?
More practice: on your own
1P(−8, −6) is on the terminal arm of θ (0° ≤ θ ≤ 360°). Find sin θ and θ to the nearest tenth of a degree.
r = √(64 + 36) = 10. sin θ = −6 ÷ 10 = −0.6. The related acute angle is sin⁻¹(0.6) ≈ 36.9°. P is in quadrant III, so θ = 180° + 36.9° = 216.9°.
2P(7, −7) is on the terminal arm of θ. Find sin θ (3 decimals) and θ.
r = √98 ≈ 9.899. sin θ ≈ −7 ÷ 9.899 ≈ −0.707. Related angle 45°, quadrant IV: θ = 360° − 45° = 315°.
3Find two angles between 0° and 360° with sin θ = 0.9. Round to the nearest tenth.
sin⁻¹(0.9) ≈ 64.2°. The other is in quadrant II: 180° − 64.2° = 115.8°.
4A seat on a wheel with radius 8 m and centre 9 m above the ground has height h = 9 + 8 sin θ. Find the height after turns of 45° and 300° (one decimal).
45°: 9 + 8(0.7071) ≈ 14.7 m. 300°: 9 + 8(−0.8660) ≈ 9 − 6.93 ≈ 2.1 m.
5Explain why sin θ can never be greater than 1.
sin θ = y ÷ r, and the height y of a point can never be more than its distance r from the origin. So y ÷ r ≤ 1. (Likewise y ≥ −r, so sin θ ≥ −1.)
Challenge
1Find all angles from 0° to 360° with sin θ = −0.4 (nearest tenth).
Related angle sin⁻¹(0.4) ≈ 23.6°. Sine is negative in quadrants III and IV: θ = 180° + 23.6° = 203.6° and θ = 360° − 23.6° = 336.4°.
2The terminal arm of θ passes through P(−5, k), and sin θ = 0.6. Find k and θ.
sin θ > 0 so k > 0. y ÷ r = 0.6 means y = 3m and r = 5m for some m > 0; then x = −4m (from x² + y² = r², x negative). −4m = −5 gives m = 1.25, so k = 3.75 (r = 6.25). θ = 180° − sin⁻¹(0.6) ≈ 180° − 36.9° = 143.1°.
3On the Ferris wheel h = 12 + 10 sin θ, find every turn angle θ between 0° and 360° where the seat is 17 m high.
17 = 12 + 10 sin θ gives sin θ = 0.5, so θ = 30° or θ = 180° − 30° = 150°.
Quiz
1. Where is the initial arm of an angle in standard position?
2. P(−6, 8) is on the terminal arm of θ. What is sin θ?
3. In which quadrant is sin θ negative and θ between 180° and 270°?
4. Which angle has the same sine as 70°?
5. On a unit circle, sin θ is
6. Which value is impossible for sin θ?
Big idea: Extending the sine ratio from right triangles to all angles from 0° to 360°, the first step of C2.3 (connecting the sine ratio to the sine function).
How to help: Use a clock face or a paper plate to show angles turning counterclockwise. Ask, "Is the point above or below the centre? Then is the sine positive or negative?"
Common difficulties:
- Calculator in radian mode: sin 30 should give 0.5.
- Thinking r can be negative. r is a distance, so it is always positive; only x and y carry signs.
- Using sin⁻¹ and stopping. The calculator gives only one angle (between −90° and 90°). Use the quadrant and the related acute angle to find the angle(s) you need.