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Grade 11 / High 2 (age 16-17)

Graphs of Trigonometric Functions

Graphs of Trigonometric Functions

A trigonometric graph is a wave that repeats the same shape forever, drawn by the height of a point circling around. That makes it ideal for anything cyclic, like a pendulum, ocean waves, or sound. The coefficient out front sets the wave height (amplitude), the number multiplying x sets how tightly it repeats (period), and an added constant shifts the graph up, down, or sideways. Drag the sliders for amplitude, period, and shift to watch the curve respond.

Basic Graphs of sin, cos, tan
💡 Key Differences
①sin x: odd, period 2π
②cos x: even, period 2π
③tan x: odd, period π, with vertical asymptotes
Amplitude & Period
1
6.28
0
0
Period & Amplitude Formulas
General Form
y = A sin(Bx + C) + D
A = amplitude, 2π/B = period, -C/B = phase shift, D = vertical shift
Period Formula
period T = |B|
Larger B → shorter period (faster oscillation)
📐 Role of Each Parameter
①A (amplitude): vertical scale
②B (angular freq): sets the period (T = 2π/|B|)
③C (phase): horizontal shift
④D (vertical shift): centerline position
Wrap-up
Trig Graph Core
y = A sin(Bx + C) + D → amplitude |A|, period |B|
Four parameters describe every sinusoid
🎯 Exam Points
①sin, cos period 2π; tan period π
②Memorize: amplitude |A|, period 2π/|B|
③Phase shift: by −C/B
④Max = A + D, min = -A + D
⑤sin and cos: cos x = sin(x + π/2)
Worked Examples & Past Exam
Example 1
Find the amplitude and period of y = 3 sin 2x.
1
In the form y = A sin(Bx), the amplitude is |A|.
amplitude = |A| = |3| = 3
2
The period is 2π/|B|.
period = |2| = π
amplitude 3, period π
Read amplitude as |A| and period as 2π/|B| directly.
Example 2
Find the maximum and minimum of y = 2 cos x + 1.
1
The range of cos x is −1 ≤ cos x ≤ 1.
-1 ≤ cos x ≤ 1
2
Substitute the endpoints to get max and min.
max = 2(1)+1 = 3, min = 2(-1)+1 = -1
max 3, min −1
Compute max = |A| + D and min = −|A| + D directly.
2022 KICE mock exam Math type, adapted
Find the period and maximum of y = 4 sin(2x − π) + 1.
period π, max 5
period 2π, max 5
period π, max 4
period π/2, max 5
period 2π, max 4
① period π, max 5
1
The period is 2π/|B| with B = 2.
period = 2 = π
2
The maximum is |A| + D.
max = |4| + 1 = 5
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