For y = a sin(bx + c) or y = a cos(bx + c), the number b controls the period, and the bracket, once factorised, controls the horizontal shift. You meet this when a question gives an equation and asks for a sketch, a maximum point or a repeat length.
This lesson builds on the graph shapes in trigonometric equations and graphs and on the roots you found in solving on a stated interval.
How do you read the period and the shift?
Treat the equation as the basic curve with three changes applied.
- Period. The basic sin x and cos x repeat every 360°. With bx inside, the period is 360°/b. The basic tan x repeats every 180°, so tan bx has period 180°/b.
- Shift. Factorise the bracket so that x stands alone: bx + c = b(x + c/b). The graph moves by c/b in the opposite direction to the sign, left if the bracket has a plus and right if it has a minus.
- Key points. Mark where the wave starts a cycle, then use quarter periods to place the maximum, midline crossing and minimum.
A period is always a positive length. Shifts are read after the factorising step, never before.
Worked example
Describe y = 3 sin(2x − 60°) + 1 and state where its first maximum occurs for x ≥ 0°.
Step 1, amplitude and midline: the 3 is the amplitude and the +1 lifts the midline to y = 1. The height runs from 1 − 3 = −2 to 1 + 3 = 4.
Step 2, period: b = 2, so the period is 360°/2 = 180°.
Step 3, shift: factorise 2x − 60° = 2(x − 30°). The curve is the basic sine shape moved 30° to the right.
Step 4, key points: the cycle starts on the midline at x = 30°, where y = 1. A quarter period is 180°/4 = 45°, so the first maximum is at x = 30° + 45° = 75° and the value is 4.
Check: substitute x = 75°: 3 sin(150° − 60°) + 1 = 3 sin 90° + 1 = 4. Also the minimum, at x = 165°: 3 sin(330° − 60°) + 1 = 3 sin 270° + 1 = −2.
The mistake to watch for
A common slip is to read the shift straight from the number in the bracket.
Mistaken answer: “Shift 60° to the right, first maximum at x = 60° + 45° = 105°.”
Test it: 3 sin(210° − 60°) + 1 = 3 sin 150° + 1 = 2.5, which is not a maximum.
The 2 is multiplying x, so the shift is 60°/2 = 30°. Factorising first is what exposes it. The substitution test in the check line is the quickest way to catch a wrong shift before you hand in the answer.
Check yourself
1. State the period and the shift of y = cos(3x + 90°).
Show answer
Period: 360°/3 = 120°. Factorise: 3x + 90° = 3(x + 30°). The shift is 30° to the left.
Period 120°, shift 30° left
2. What is the period of y = 2 sin(x/2), and how many complete waves appear between 0° and 720°?
Show answer
Here b = 1/2, so the period is 360° ÷ 1/2 = 720°. Exactly one complete wave fits between 0° and 720°.
Period 720°, one wave
3. For y = 4 cos(2x − π/3), find the period and the smallest positive x where y is a maximum, in radians.
Show answer
Period: 2π/2 = π. Factorise: 2x − π/3 = 2(x − π/6), so the shift is π/6 to the right. The cosine is at its maximum when its bracket is 0, so x = π/6, and y = 4.
Period π, maximum at x = π/6
Where this leads next
Now use the period when solving harder equations in finding all roots after a variable substitution. The quadratic structure explorer shows how a graph’s key points connect to its equation, which is the same habit of matching a picture to numbers. The non-calculator working trainer helps with the exact values you need for key points.
Shift and period questions get easier once someone checks your sketch step by step, which is what happens in online one-to-one Additional Mathematics tuition.