Sine, cosine, tangent, and all six trig functions — how to graph them, transform them, and read equations from them.
| x | sin x | Description |
|---|---|---|
| 0 | 0 | Start (zero crossing) |
| π/2 | 1 | Maximum |
| π | 0 | Middle zero crossing |
| 3π/2 | −1 | Minimum |
| 2π | 0 | End (zero crossing) |
Graph shape mnemonic
Sine starts at zero, rises to 1, falls back through zero, hits −1, and returns to zero. Think: "starts flat, peaks up, dips down". It crosses the x-axis at every multiple of π.
| x | cos x | Description |
|---|---|---|
| 0 | 1 | Start (maximum) |
| π/2 | 0 | Zero crossing (falling) |
| π | −1 | Minimum |
| 3π/2 | 0 | Zero crossing (rising) |
| 2π | 1 | End (maximum) |
Phase relationship: cos x = sin(x + π/2)
The cosine graph is the sine graph shifted π/2 units to the LEFT. Both have the same amplitude and period — cosine just starts at its maximum (1) rather than at zero. This relationship is useful when converting between sin and cos forms.
Same formula applies to cosine. Each letter controls exactly one transformation.
Amplitude
|A|
Vertical stretch/compression. If |A| > 1 the graph is taller; if |A| < 1 it is shorter. A negative A reflects the graph over the midline (flips it upside down). The maximum is D + |A| and the minimum is D − |A|.
y = 3 sin x → amplitude = 3, max = 3, min = −3
Period
2π / |B|
Horizontal stretch/compression. Larger |B| compresses the graph (shorter period). Smaller |B| stretches it (longer period). B does NOT affect amplitude or vertical position.
y = sin(2x) → period = 2π/2 = π
Phase Shift
−C / B
Horizontal translation. Positive result → shift right; negative → shift left. Warning: the sign is easy to flip. In y = sin(x − π/4), C = −π/4, so phase shift = −(−π/4)/1 = π/4 right.
y = sin(x + π/6) → phase shift = −π/6 (left)
Vertical Shift
D (the midline)
Moves the entire graph up (D > 0) or down (D < 0). The midline of the graph is y = D. Max = D + |A|, Min = D − |A|.
y = sin x + 2 → midline y = 2, max = 3, min = 1
tan(−π/4) = −1
tan(0) = 0 (x-intercept)
tan(π/4) = 1
Vertical Asymptotes
tan x is undefined wherever cos x = 0, because tan x = sin x / cos x. That happens at x = ±π/2, ±3π/2, … or generally x = π/2 + nπ for any integer n. The graph shoots to ±∞ at each asymptote. For y = tan(Bx), asymptotes occur at x = π/(2B) + nπ/B.
Each of these is derived from a basic function by taking its reciprocal. Sketch the basic function first, then use it as a guide.
Period
π
Asymptotes
x = nπ (where sin x = 0)
Decreasing curve through each period. Crosses x-axis at x = π/2 + nπ. Asymptotes where tan x has x-intercepts and vice versa — the two graphs swap roles.
Cot is the tan graph flipped horizontally and shifted π/2.
Period
2π
Asymptotes
x = π/2 + nπ (where cos x = 0)
U-shaped arches that open up at each cosine maximum and down at each minimum. The vertices of the arches touch the cosine graph. Draw cosine first, then draw arches through each max and min.
Sec and cos share maxima/minima at the same x values.
Period
2π
Asymptotes
x = nπ (where sin x = 0)
U-shaped arches opening up from each sine maximum and down from each minimum. Asymptotes at every multiple of π. Draw the sine curve first as a guide.
Csc arches sit on top of and below the sine wave.
Extract A, B, C, D
Rewrite the function as y = A sin(Bx + C) + D. If written as y = A sin(B(x − h)) + D, identify h directly as the phase shift and note that C = −Bh.
Calculate the four parameters
Amplitude = |A|. Period = 2π/|B|. Phase shift = −C/B (positive = right, negative = left). Vertical shift = D (this is also the midline y = D).
Draw the midline
Draw a dashed horizontal line at y = D. The graph oscillates symmetrically above and below this line. Mark D + |A| (max) and D − |A| (min) on the y-axis.
Find and plot the 5 key x-values
Start at the phase shift x₀ = −C/B. Space 4 points equally at intervals of period/4. For sine: the pattern at these 5 points is midline → max → midline → min → midline. For cosine: max → midline → min → midline → max.
Sketch the curve and extend
Connect the 5 key points with a smooth sinusoidal curve. Remember: if A < 0 the pattern is inverted (midline → min → midline → max → midline for negative sine). Extend the pattern left and right as needed.
Example 1 — Basic sine graph
y = sin x
Amplitude = 1, Period = 2π, Phase shift = 0, Vertical shift = 0.
Key points: (0, 0) → (π/2, 1) → (π, 0) → (3π/2, −1) → (2π, 0).
Graph: smooth S-curve starting at origin, peaking at π/2, returning to zero at π.
Example 2 — Fully transformed cosine
y = −3 cos(2x − π) + 1
Identify: A = −3, B = 2, C = −π, D = 1
Amplitude = |−3| = 3
Period = 2π / 2 = π
Phase shift = −(−π)/2 = π/2 to the right
Vertical shift = 1 (midline: y = 1)
Max = 1 + 3 = 4, Min = 1 − 3 = −2
A is negative → cosine pattern inverted: start at min (not max)
Key x-values (spacing = period/4 = π/4):
x = π/2 → y = −2 (inverted start: min)
x = π/2 + π/4 = 3π/4 → y = 1 (midline)
x = π → y = 4 (max)
x = 5π/4 → y = 1 (midline)
x = 3π/2 → y = −2 (min again)
Smooth curve through these 5 points, period π.
Example 3 — Period change only
y = sin(πx)
A = 1, B = π, C = 0, D = 0
Period = 2π / π = 2
Key points: (0, 0) → (0.5, 1) → (1, 0) → (1.5, −1) → (2, 0)
One full wave fits between x = 0 and x = 2 (not 0 to 2π).
Example 4 — Finding the equation from a graph
A graph has: maximum = 5, minimum = −1, period = 4π, first maximum at x = π/2.
Step 1: Amplitude A = (max − min)/2 = (5 − (−1))/2 = 3
Step 2: Vertical shift D = (max + min)/2 = (5 + (−1))/2 = 2
Step 3: Period = 4π → B = 2π/4π = 1/2
Step 4: Phase shift. Cosine normally has its first max at x = 0. This graph has first max at x = π/2 → phase shift = π/2 to the right.
Using phase shift = −C/B: π/2 = −C/(1/2) → C = −π/4
Equation: y = 3 cos((1/2)x − π/4) + 2
Verify: at x = π/2, inside = (1/2)(π/2) − π/4 = π/4 − π/4 = 0, cos(0) = 1 → y = 3(1) + 2 = 5 ✓
Example 5 — Graphing a transformed tangent
y = 2 tan(x/2)
A = 2, B = 1/2, C = 0, D = 0
Period = π / (1/2) = 2π
Asymptotes: x = π/2 ÷ (1/2) = ±π, then ±3π, ±5π, …
At x = 0: y = 0 (x-intercept)
At x = π/2: y = 2 tan(π/4) = 2(1) = 2
At x = −π/2: y = 2 tan(−π/4) = −2
Stretch vertically by 2, period doubled to 2π, asymptotes at ±π.
Using 2π/B without the absolute value
Period = 2π / |B|. If B is negative (e.g., y = sin(−2x)), |B| = 2 and period = π. Ignoring the absolute value gives a negative period, which is meaningless.
Getting the phase shift sign wrong
Phase shift = −C/B. In y = sin(x + π/3), C = +π/3, so phase shift = −π/3 (LEFT, not right). The plus inside the parentheses means LEFT. Many students see '+' and think right.
Forgetting to factor out B before reading phase shift
In y = sin(2x + π), you cannot read phase shift as π. Factor: y = sin(2(x + π/2)), so the phase shift is π/2 left. If you skip factoring, you get the wrong answer by a factor of B.
Thinking tangent has period 2π
Tangent period is π, not 2π. Only sine and cosine have period 2π. Cotangent also has period π. Secant and cosecant have period 2π.
Forgetting that a negative A flips the graph, not the period or phase
A = −2 means amplitude 2 with a reflection over the midline. The period and phase shift are unchanged. Flip the key point pattern: for negative sine, start at midline going DOWN instead of up.
| Function | Period | Amplitude | Range | Asymptotes at |
|---|---|---|---|---|
| sin x | 2π | 1 | [−1, 1] | none |
| cos x | 2π | 1 | [−1, 1] | none |
| tan x | π | none | (−∞, ∞) | π/2 + nπ |
| cot x | π | none | (−∞, ∞) | nπ |
| sec x | 2π | none | (−∞,−1] ∪ [1,∞) | π/2 + nπ |
| csc x | 2π | none | (−∞,−1] ∪ [1,∞) | nπ |
The basic sine and cosine functions both have a period of 2π (approximately 6.28). This means the graph repeats every 2π units along the x-axis. For a transformed function y = A sin(Bx + C) + D, the period becomes 2π/|B|. If B = 2, the period is π; if B = 1/2, the period is 4π.
For y = A sin(Bx + C) + D, the phase shift is −C/B. The negative sign is critical: if C is positive, the shift is to the LEFT. Example: y = sin(x − π/3) has B = 1 and C = −π/3, so phase shift = −(−π/3)/1 = π/3 to the right. Always factor out B first if the function is written as y = sin(B(x − h)) + D, in which case h is directly the phase shift.
The basic tangent function has a period of π, not 2π. This is because tangent completes one full cycle between consecutive vertical asymptotes, which are spaced π apart. For y = A tan(Bx + C) + D, the period is π/|B|.
Use the 5-step process: (1) Identify A, B, C, D from y = A sin(Bx + C) + D. (2) Calculate amplitude |A|, period 2π/|B|, phase shift −C/B, vertical shift D. (3) Find the midline y = D. (4) Determine the 5 key points by starting at the phase shift and spacing them period/4 apart. (5) Apply amplitude and vertical shift: max = D + |A|, min = D − |A|, midline crossings = D.
Read four values from the graph: (1) Amplitude A = (max − min)/2. (2) Vertical shift D = (max + min)/2 (the midline). (3) Period from graph, then B = 2π/period. (4) Phase shift: find where the first maximum or zero crossing occurs and use that to solve for C in the formula phase shift = −C/B, giving C = −B × (phase shift). Assemble y = A sin(Bx + C) + D (or cosine, depending on which function fits the starting behavior).
Unit circle, right triangle trig, all six functions
Pythagorean, double angle, half angle, sum/difference formulas
Finding all solutions on [0, 2π) and the general solution
arcsin, arccos, arctan and their graphs and domains
The general theory of shifting, stretching, reflecting
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Interactive graphing problems with step-by-step solutions — identify amplitude, period, phase shift, and write equations from graphs.
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