1
The graph of the sine function: a table of values, domain, range, period 2π, maximum and minimum
Mathematics Sec 1 Term 1, Unit 4, Lessons 5–6
GoalDraw y = sin θ and read its domain, range, period, maximum and minimum.
New words
period · the length of one full cycle before the graph repeatsrange · all the values y can takemaximum and minimum · the highest and the lowest value of the function
Explanation
Plot the points (θ, sin θ) from the table, with θ in radians, and join them smoothly. The result is a wave. The domain is all real numbers (any angle). The range is [−1, 1] because sin θ = y on the unit circle. The wave rises to a maximum of 1 at θ = π/2, falls through 0 at π, reaches a minimum of −1 at 3π/2 and returns to 0 at 2π, where it starts again. So the period is 2π: sin(θ + 2π) = sin θ. The graph passes through the origin and is symmetrical about it, because sin(−θ) = −sin θ. It cuts the θ-axis at 0, π, 2π, … that is at every multiple of π.
Worked examples
| θ | 0 | π/6 | π/2 | 5π/6 | π | 7π/6 | 3π/2 | 11π/6 | 2π |
|---|---|---|---|---|---|---|---|---|---|
| sin θ | 0 | 1/2 | 1 | 1/2 | 0 | −1/2 | −1 | −1/2 | 0 |
Where is sin θ = 0 for 0 ≤ θ ≤ 3π? At θ = 0, π, 2π and 3π (every multiple of π).
Find sin(13π/2). 13π/2 = 6π + π/2, and 6π is three full periods, so sin(13π/2) = sin(π/2) = 1.
Class activity
Draw it by hand. On squared paper draw the sine wave from 0 to 2π using the table, taking 3 cm for π. Mark the maximum and minimum points, then continue the pattern for another period.
Practice
1
State the domain, range and period of y = sin θ.
Domain: all real numbers. Range: [−1, 1]. Period: 2π.
2
At which θ in [0, 2π] is sin θ = 1? And −1?
θ = π/2 for 1 and θ = 3π/2 for −1.
3
Find sin(5π/2) and sin(−π/2).
sin(5π/2) = sin(π/2) = 1; sin(−π/2) = −1.
4
For which θ in [0, 2π] is sin θ positive?
For 0 < θ < π (above the axis).
5
Is sin(θ + 4π) equal to sin θ? Why?
Yes: 4π is two whole periods.