Trigonometric Graphs
Higher
Sine, cosine and tangent graphs show how trigonometric values change as an angle changes. Learn their shapes, repeat patterns and how to find every graphical solution within an interval.
Before you start: Right-Angled Trigonometry and Exact Values, Linear Graphs and Functions.
1. Inputs are angles, outputs are numbers
In y = sin x, the input x is an angle in degrees and the output y is its sine. The same idea applies to cosine and tangent. Use degree mode on your calculator. The horizontal axis is an angle scale, not a triangle side.
A period is the smallest positive horizontal interval after which the whole graph repeats. Sine and cosine have period 360°; tangent has period 180°. The graphs extend to negative angles and beyond 360°.
Plot key points on evenly spaced angle axes, adding intermediate values to shape a smooth curve. A graph on 0° to 360° is one useful viewing window, not the full domain. Triangle ratios and acute-angle exact values are taught in Right-Angled Trigonometry and Exact Values.
2. Sine and cosine
Worked example — construct one full cycle
| x (degrees) | sin x | cos x |
|---|---|---|
| 0 | 0 | 1 |
| 90 | 1 | 0 |
| 180 | 0 | -1 |
| 270 | -1 | 0 |
| 360 | 0 | 1 |
For sine, start at (0°, 0), rise to (90°, 1), fall through (180°, 0) to (270°, −1), then rise to (360°, 0). For cosine, start at (0°, 1), fall through (90°, 0) to (180°, −1), then rise through (270°, 0) to (360°, 1).
Both graphs have values from −1 to 1 inclusive. Their maximum is 1 and minimum is −1, but they reach them at different inputs. On 0° ≤ x ≤ 360°, sine is zero at 0°, 180°, 360°; cosine is zero at 90°, 270°.
Worked example — read all solutions in an interval
To solve sin x = ½ for 0° ≤ x ≤ 360°, draw y = ½. It meets the sine curve twice: x = 30° and 150°. Both heights are ½. For cos x = −½, use y = −½ on the cosine graph: x = 120° and 240°.
A calculator’s inverse-sine result of 30° gives one angle. It does not automatically list every crossing in your interval. Use the graph’s symmetry and repetition, and include endpoints only when they satisfy the equation and the interval.
3. Tangent: repeating branches and gaps
The tangent graph passes through (0°, 0), (45°, 1), (135°, −1), (180°, 0), (225°, 1), (315°, −1) and (360°, 0). It increases along each branch and has no maximum or minimum.
tan 90° and tan 270° are undefined. Draw separate branches on each side of these inputs. The vertical lines x = 90° and x = 270° are asymptotes: the values grow in magnitude near them, but there is no plotted point at either angle.
Worked example — solve tan x = 1
On 0° ≤ x ≤ 360°, the line y = 1 crosses at x = 45° and 225°. The gap of 180° matches the tangent period. Do not include 90° or 270°: an asymptote is not an extra solution.
Outside this window the undefined inputs continue every 180°: …, −90°, 90°, 270°, 450°, … . Zeros also repeat every 180°: …, −180°, 0°, 180°, 360°, … . The output is not restricted to −1 to 1.
4. Estimates, negative angles and repeated periods
Worked example — sin x = 0.7
Read x ≈ 44° and 136° for 0° ≤ x ≤ 360°. A hand-drawn reading within about 2° is sensible on this scale. The line y = 0.7 is above the axis; sine is below the axis between 180° and 360°, so there are no further crossings there.
Worked example — move beyond one cycle
sin 390° = sin 30° = ½ because 390° is one full 360° turn after 30°. Also sin(−30°) = −½, as the sine curve just left of the origin is below the axis. cos(−60°) = ½; cosine has reflection symmetry in the y-axis. tan(−135°) = tan 45° = 1 because the inputs differ by 180°.
Worked example — solve on a wider interval
For sin x = ½ on −360° ≤ x ≤ 360°, start with 30° and 150°. Subtract one period, 360°, to get −330° and −210°. All four lie in the interval: −330°, −210°, 30°, 150°. Adding another period or subtracting another would leave it.
A reliable route is: identify the height → find every crossing in one cycle → repeat using the correct period → keep only inputs in the stated interval.
Think Like a Mathematician
Why can sin x = 1 have fewer solutions than sin x = ½ in one cycle? The line y = 1 touches the top of the sine wave once, while y = ½ crosses it twice. On 0° ≤ x ≤ 360°, sin x = 0 has three listed answers because both endpoints are included. Counting solutions always depends on both the height and the interval.
Common Mistakes
Wrong calculator mode: use degrees. Wrong period: tangent repeats after 180°, not 360°. Joined gaps: never draw through 90° or 270° on the tangent graph. Missing answers: the principal inverse-trig value may be only one crossing. Impossible height: sin x = 1.2 has no real solution.
Practice — Try It Yourself
Work on squared paper where a graph is requested. Try each question before opening its answer. Graph readings are estimates: give only the precision your scale supports.
1. State the period and output range of sine and cosine. How does tangent differ?
Sine and cosine: period 360°, range −1 ≤ y ≤ 1. Tangent: period 180°, can take any real output and has undefined inputs at 90° plus whole multiples of 180°.
2. Sketch y = cos x for 0° ≤ x ≤ 360°, labelling its zeros, minimum and maxima.
Plot (0°, 1), (90°, 0), (180°, −1), (270°, 0), (360°, 1), then draw a smooth wave. Zeros: 90°, 270°. Minimum: (180°, −1). Maxima in the displayed interval: (0°, 1), (360°, 1). Compare with the worked cosine graph.
3. Read sin 270°, cos 180° and tan 225°. Is tan 90° equal to zero?
sin 270° = −1; cos 180° = −1; tan 225° = 1. tan 90° is undefined, not zero.
4. Solve sin x = −½ for 0° ≤ x ≤ 360°.
Use the horizontal height −½ on the lower half of the sine wave: x = 210° and 330°. These are 30° after 180° and 30° before 360°.
5. Solve cos x = 0 for −180° ≤ x ≤ 180°. Give every solution.
x = −90° and 90°. The endpoints −180° and 180° both have cosine −1, so they are not solutions.
6. Solve tan x = −1 for 0° ≤ x ≤ 360°.
The crossings are 135° and 315°, separated by 180°. A calculator may return −45°, which has the right tangent value but lies outside this interval.
7. Use the sine graph to estimate all solutions of sin x = 0.7 for 0° ≤ x ≤ 360°. Explain why giving only 44° is incomplete.
x ≈ 44° and 136°. The horizontal line meets both sides of the upper sine arch; the second reading is 180° − 44° ≈ 136°. There are no crossings on the negative half-cycle.
8. Find sin 450°, cos(−360°) and tan 405° using periodicity.
sin 450° = sin 90° = 1. cos(−360°) = cos 0° = 1. tan 405° = tan 45° = 1 (subtract 360°, two tangent periods).
9. Solve sin x = 0 for −180° ≤ x ≤ 360°. Are the endpoints included?
−180°, 0°, 180°, 360°. All are zeros and both endpoints are included because the interval uses ≤. Changing an endpoint to a strict inequality would exclude that endpoint.
Frequently Asked Questions
Are the sine and cosine graphs only defined from 0° to 360°?
No. They extend in both directions and repeat every 360°. A question’s interval limits which solutions you report.
Why is tangent drawn in pieces?
It is undefined at certain angles. The separate branches approach vertical asymptotes; drawing across them would invent values that do not exist.
Does this replace triangle trigonometry?
No. Use Right-Angled Trigonometry and Exact Values for right-angled triangles. Here the focus is on the functions and their graphs.
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