1The idea in one paragraph

As a point runs round the unit circle, its height, sin θ, rises to 1, falls through 0 to −1 and comes back; plot that height against the angle and you get a wave that repeats every 2π. Its horizontal position, cos θ, gives the same wave shifted a quarter of a turn. These are the circular functions. Every other wave on the syllabus is one of them stretched and moved: in f(x) = a sin(b(x + c)) + d, a sets the height of the wave, b how many waves fit into 2π, c slides it sideways and d lifts it. Tan x is different, the gradient of the arm, which blows up to infinity every time the arm is vertical, so its graph is a repeating curve broken by asymptotes. Anything that repeats regularly, a tide, a turning wheel, the length of a day, can be modelled with one of these waves.

2From the circle to the graph

A circular function takes an angle and returns a coordinate of the point on the unit circle. Figure 1 unrolls the circle: the height of P in panel (a) becomes the y-value of the graph in panel (b) at horizontal position θ.

Follow P once round. It starts at height 0, climbs to 1 at π/2, returns to 0 at π, reaches −1 at 3π/2 and gets back to 0 at 2π. Then it goes round again, and the graph repeats. A function that repeats is periodic, and the length of one repeat is its period.

On examination papers the x-axis is in radians unless the question says degrees. A graph in degrees is the same shape with 360° where the radian graph has 2π.

3The three basic graphs

Figure 2 draws y = sin x and y = cos x together.

Three features come straight from 3.5.

• Cosine is sine moved π/2 to the left, because cos x = sin(x + π/2). A cosine wave starts at its maximum; a sine wave starts on its middle line, going up.
• Cosine is symmetric in the y-axis, because cos(−x) = cos x. Sine has half-turn symmetry about the origin, because sin(−x) = −sin x.
• Tangent has period π, not 2π, because tan(x + π) = tan x: the arm and its half-turn have the same gradient.

The graph of y = tan x in Figure 3(a) passes through the origin with gradient 1, climbs without limit towards the vertical asymptote x = π/2 (where cos x = 0 and the tangent is undefined), and reappears from −∞ on the other side. Each branch is a copy of the one before, moved π to the right.

4Amplitude, period and the principal axis

Three words describe any sine or cosine wave.

• The principal axis is the horizontal line through the middle of the wave: y = (max + min) ÷ 2.
• The amplitude is the distance from the principal axis to a maximum: (max − min) ÷ 2. It is a distance, so it is never negative.
• The period is the horizontal length of one complete wave.

For the general wave,

f(x) = a sin(b(x + c)) + d has amplitude |a|, period 2π ÷ |b| (or 360° ÷ |b| in degrees), horizontal translation c to the left (that is, −c), and principal axis y = d. Its range is d − |a| ≤ f(x) ≤ d + |a|.

The same is true with cos in place of sin. For tan(b(x + c)) + d there is no amplitude and the period is π ÷ |b|.

Why 2π ÷ b? Inside the bracket, the sine completes one wave when b(x + c) grows by 2π, and that takes an increase in x of 2π ÷ b. So a larger b squeezes more waves into the same space: sin 3x has period 2π/3, three waves in every 2π.

Figure 4 reads every number off one example, y = 3 sin(2(x − π/6)) + 1.

A negative a reflects the wave in its principal axis: y = −2 sin x starts at 0 and goes down first. The amplitude is still 2.

Degrees. f(x) = 2 sin(4x) + 1 with 0° ≤ x ≤ 180° has period 360° ÷ 4 = 90°, so two complete waves fit in the domain, with maxima of 3 at x = 22.5° and 112.5°.

Tangent. In Figure 3(b), y = tan(x − π/4) is y = tan x moved π/4 to the right. Every zero and every asymptote moves with it: zeros at π/4 and 5π/4, asymptotes at 3π/4 and 7π/4. And y = tan 2x has period π/2, with asymptotes at π/4, 3π/4, 5π/4 and 7π/4 in 0 to 2π.

5Transformations, one at a time

This is 2.11 applied to one family. Going from y = sin x to y = a sin(b(x + c)) + d takes four transformations, and Figure 5 draws them in order.

The bracket matters. The translation is only c when b has been factored out. y = sin(2x + π/3) is not a shift of π/3: factorise first, sin(2(x + π/6)), and the shift is π/6 to the left. When the stretch comes before the translation, as here, the translation is by c; if you do the translation first you must translate by the unfactorised amount and then stretch. The safe habit is always to factorise, then state the stretch and the translation.

Worked example 1. Describe the transformations that map y = cos x onto y = 4 cos(x/2) − 1, and state the amplitude, period and range.

"Describe fully" means naming the type of transformation, its direction and its size. "A stretch by 2" without saying horizontal, or "moved down" without the distance, loses the mark.

Worked example 2 (sketching, Paper 1). To sketch y = 3 sin(2(x − π/6)) + 1 on 0 ≤ x ≤ 2π, as in Figure 4:

1. Draw the principal axis y = 1 lightly, and the lines y = 4 and y = −2.
2. The wave starts on its principal axis, rising, where the bracket is 0: at x = π/6.
3. Mark the quarter-periods from there, π/4 apart: principal axis at π/6, maximum at 5π/12, principal axis at 2π/3, minimum at 11π/12, principal axis at 7π/6.
4. Repeat for the second period, then extend the curve back to x = 0 and on to x = 2π, and work out the end values: at x = 0, y = 3 sin(−π/3) + 1 = 1 − 3√3/2 ≈ −1.60.

An examiner looks for the right number of waves in the domain, the correct maximum and minimum values, the right x-intercept or turning-point positions, and the endpoints of the domain.

6Finding the equation from a graph

Often the graph, or two of its points, comes first and you find the function. Use the maximum M and the minimum m.

Worked example 3. A wave has a maximum at (1, 7) and the next minimum at (4, 1), as in Figure 6. Find an equation in the form y = a cos(b(x − h)) + d, and one in the form y = a sin(b(x + c)) + d.

Check either one at x = 4: 3 cos(π) + 4 = 1 ✓, and 3 sin(3π/2) + 4 = 1 ✓. Both are right. The equation of a wave is not unique: you can shift by any whole number of periods, or use −cos, and still describe the same graph. The examiner accepts any correct form unless the question fixes one.

7Real-life contexts

A circular-function model fits anything that rises and falls with a fixed period: tides, a seat on a turning wheel, hours of daylight over a year, the temperature over a day. The method is always the one in section 6: get the maximum, the minimum and the period from the story, then choose where the wave starts.

Worked example 4 (tide, Paper 2). In an invented harbour, high water of 7.8 m is at 03:00 and the next low water, 1.4 m, is at 09:15. The depth d metres, t hours after midnight, is modelled by d(t) = a cos(b(t − c)) + p.

The depth at noon is d(12) = 3.2 cos(1.44π) + 4.6 = 4.00 m (3 s.f.). A boat needs at least 5 m of water. On the GDC, graph y = d(t) and y = 5 for 0 ≤ t ≤ 24 and find the intersections: t = 0.124, 5.88, 12.6 and 18.4. So the boat can use the harbour from about 00:07 to 05:53 and from 12:37 to 18:23, two windows of 5 h 45 min each. Figure 7 shades them.

Worked example 5 (a wheel, Paper 2). An invented observation wheel has radius 20 m and its centre 23 m above the ground. It turns once every 12 minutes, and a rider boards at the lowest point at t = 0. Find the height h(t) and the time per turn the rider is above 35 m.

The rider starts at the minimum, 23 − 20 = 3 m, and the maximum is 43 m. A cosine wave starts at a maximum, so a negative cosine starts at a minimum:

Figure 8 shows the wheel and the graph. Check that the model makes physical sense: h(0) = 3, h(6) = 43 at the top after half a turn, and h(12) = 3 again.

Regression on a GDC. Given a table of data, a GDC's sinusoidal regression finds the wave of best fit. The guide warns that not all technology gives it in the syllabus form. Many calculators return y = a sin(bx + c) + d, where c is not the translation. Factorise to convert: y = 2.1 sin(0.52x − 1.3) + 5 is y = 2.1 sin(0.52(x − 2.5)) + 5, because 1.3 ÷ 0.52 = 2.5, so the wave is translated 2.5 to the right. Always write out which form your calculator used before you interpret c.