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Mathematics · High School · Trigonometry

Graphs of trigonometric functions

The idea

Graphed against time, sine and cosine become waves — and waves run the physical world, from sound and tides to alternating current and daylight hours. The general form y = A sin(B(x − C)) + D packs four dials: |A| is the amplitude (half the wave's total height), 2π/B is the period (the length of one full cycle), C is the phase shift (horizontal), and D sets the midline (vertical). Each dial is one of the function transformations you already use, applied to a periodic parent graph.

Read a formula in this order: midline first, then amplitude (maximum = D + |A|, minimum = D − |A|), then period, then shift — and watch for the one trap that catches nearly everyone. The phase shift can be read only AFTER factoring B out of the argument: in sin(2x − π/3) the shift is not π/3, because rewriting the argument as 2(x − π/6) reveals the true shift of π/6. When sketching, plot the midline crossings and the extremes — five key points per cycle — and the curve practically draws itself.

Worked example

For y = 3 sin(2x − π/3) + 1, find the amplitude, period, phase shift, and midline, then find the coordinates of the first maximum to the right of x = 0.

  1. Amplitude is |A| = 3 and the midline is y = 1, so the wave oscillates between 1 − 3 = −2 and 1 + 3 = 4.
  2. B = 2 gives period 2π/2 = π: the wave repeats every π units, twice as often as plain sine.
  3. Factor the argument before reading the shift: 2x − π/3 = 2(x − π/6), so the graph is shifted π/6 to the right — not π/3.
  4. A sine curve peaks where its argument equals π/2, so set 2x − π/3 = π/2. Then 2x = π/2 + π/3 = 5π/6, giving x = 5π/12, where y reaches the maximum 1 + 3 = 4.
  5. Check: at x = 5π/12 the argument is 5π/6 − 2π/6 = π/2 and sin(π/2) = 1, so y = 3 × 1 + 1 = 4. The point (5π/12, 4) also sits within the first period, as expected.

Answer. Amplitude 3, period π, phase shift π/6 right, midline y = 1; the first maximum right of 0 is at (5π/12, 4).

Check your understanding

  • Why is the period 2π/B rather than B itself, and what does doubling B do to the wave physically?
  • How can you read the maximum and minimum values of the graph straight from A and D without plotting anything?
  • Why must B be factored out of the argument before reading the phase shift, and what error results from skipping that step?
  • How would you choose A, B, C, and D to model a real periodic quantity, like the hours of daylight across a year?

Build the foundations first

Graphs of trigonometric functions builds on these concepts. If any feel shaky, start there.

Functions (intro)Angle relationshipsThe coordinate plane (intro)
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