KCSE 2025 Maths P2 Q16 — Trigonometric Graph: Amplitude and Period

KCSE 2025 Form 4 Trigonometry

Published

The Question

“The graph shown represents the trigonometric function y = A sin(omega x + 30 degrees) for x between -90 degrees and 90 degrees. The curve oscillates between a maximum of y = 2 and a minimum of y = -2, crosses the y-axis at y = 1, and has neighbouring peaks at x = -75 degrees and x = +15 degrees. Determine the values of the scalars A and omega.”

1

Read the amplitude to find A

The amplitude of a sine curve is the greatest height it reaches above the middle line, and for a function of the form A sin(...) that height is exactly the value of A. Read the top and bottom of the wave off the graph: the peaks sit at y = 2 and the troughs at y = -2, so the amplitude is 2. That gives A directly.

amplitude=A=2\text{amplitude} = A = 2
2

Confirm A using the y-intercept

Check the value of A against a known point on the curve. The graph crosses the y-axis at y = 1, which is where x = 0. Substituting x = 0 into the function should reproduce that height, and it does, so A = 2 is correct.

y=2sin⁡(0+30∘)=2sin⁡30∘=2×12=1y = 2\sin(0 + 30^\circ) = 2\sin 30^\circ = 2 \times \tfrac{1}{2} = 1
3

Find the period from two peaks

The scalar omega controls how stretched or squashed the wave is, and it is found from the period. The period is the horizontal distance between two neighbouring peaks. One peak is at x = -75 degrees and the next at x = +15 degrees, so subtract to get the period.

T=15∘−(−75∘)=90∘T = 15^\circ - (-75^\circ) = 90^\circ
4

Solve for omega

For a sine function the period and omega are linked by the rule that the period equals 360 degrees divided by omega. Rearrange to make omega the subject, then substitute the period of 90 degrees to find its value.

T=360∘ω⇒ω=360∘TT = \frac{360^\circ}{\omega} \Rightarrow \omega = \frac{360^\circ}{T}
ω=360∘90∘=4\omega = \frac{360^\circ}{90^\circ} = 4

Final Result

The scalars are A = 2 and omega = 4, so the function is y = 2 sin(4x + 30 degrees).

Why this method works

The amplitude method works because sine itself only ever ranges from -1 to +1, so multiplying by A stretches that range to run from -A to +A; the peak height is therefore A. Omega compresses the wave horizontally: a full sine cycle naturally spans 360 degrees, but replacing x by omega x makes the argument reach 360 degrees after only 360/omega degrees of x, which is exactly the period. Measuring the real period off the graph and inverting that relationship recovers omega.

✓

With A = 2 and omega = 4, the y-intercept is 2 sin(30) = 1 and the period is 360/4 = 90 degrees, both matching the graph.