Transforming trigonometric graphs

For y = sin(bx), one full sine cycle occurs when bx changes by 360°. With b ≠ 0:

sin 2x is compressed horizontally by factor 1/2 and has period 180°. cos(x/2) is stretched horizontally by factor 2 and has period 720°. A negative b also reverses the horizontal direction.

If a or b is zero, an expression a f(bx) + d may be constant (provided f is defined there); the usual non-constant fundamental-period formula no longer applies.

In y = sin(x − 60°), a familiar sine input t occurs at x = t + 60°. The graph shifts 60° right. In y = tan(x + 30°), it shifts 30° left.

tan(x + 30°) zeros: x = −30° + 180°n
asymptotes: x = 60° + 180°n

For a combined expression, factor the coefficient of x before reading the shift. sin(2x − 60°) = sin(2(x − 30°)) shifts the compressed graph 30° right, not 60°.

Consider y = 1 + 2cos(2(x − 30°)). Starting from a point (t, cos t), solve t = 2(x − 30°).

(0°,1) → (30°,3)
(90°,0) → (75°,1)
(180°,−1) → (120°,−1)
(270°,0) → (165°,1)
(360°,1) → (210°,3)

The amplitude is 2, midline 1 and period 180°. These mapped points are on the midline or extrema; midline crossings are not necessarily x-axis crossings.

cos(−x) = cos x
sin(−x) = −sin x
tan(−x) = −tan x

The unit-circle coordinates also give the quarter-turn and complement relations:

cos(x − 90°) = sin x
sin(x − 90°) = −cos x
sin(90° − x) = cos x
cos(90° − x) = sin x

If y = sin(px) has its first positive zero at 45°, then 180°/|p| = 45°, so |p| = 4. Both p = 4 and p = −4 have those zeros. A positive p assumption or the direction of the crossing is needed to choose the sign.

For y = sin(x + k), a rising zero at x = −40° gives k = 40° + 360°n. The smallest positive choice is 40°, but every integer n gives the same graph.

Distinguish a rising zero from a falling one: both are zeros, but their phase differs by 180°. A graph’s maximum and minimum determine its midline and amplitude by their average and half their difference.

A simplified water-level model is h(t) = 2 + sin(45t)°, where h is metres above a fixed datum and t is hours after 09:00. Its period is 360/45 = 8 hours, range 1–3 m and first maximum is at t = 2.

During 0 ≤ t ≤ 8, the level is at least 2.5 m when sin(45t)° ≥ 1/2. Thus 30° ≤ 45t ≤ 150°, giving 2/3 ≤ t ≤ 10/3: from 09:40 to 12:20.

Real water levels need not repeat perfectly. The model ignores weather and changing tides; use it only over a justified time range. In a spatial model y = 1 + 0.4sin(90x)°, with x in metres, one cycle is 4 m, so an interval of length 16 m contains four complete cycles.

To solve an equation graphically, write it as two functions and count their intersections on the specified interval.

For tan x = 2cos x on 0° ≤ x ≤ 180°, the branch on 0° ≤ x < 90° rises from 0 towards infinity while 2cos x falls from 2 towards 0. They cross once. On 90° < x ≤ 180°, tangent rises from negative infinity to 0 while 2cos x falls from 0 to −2, so they cross once again.

There are two solutions. x = 90° is excluded because tangent is undefined there, not an extra solution. A coarse plot joined across that gap could create a false intersection.