The six graphs and their periods.

The six trigonometric graphs with dashed asymptotes; a ghost copy of the chosen graph slides sideways and locks green when it overlays the original, revealing the period.

Choose tan and find the smallest non-zero shift that makes the ghost overlay the graph (π, not 2π).

Stating the period of tan x as 2π; giving the range of sec x as [−1, 1].

• sin x and cos x are defined for every real x, and their range is [−1, 1].
• tan x has domain R − {(2n + 1)π/2 : n ∈ Z} and range R; cot x has domain R − {nπ : n ∈ Z} and range R.
• sec x has domain R − {(2n + 1)π/2 : n ∈ Z}, and cosec x has domain R − {nπ : n ∈ Z}; both have range (−∞, −1] ∪ [1, ∞), because |sec x| = 1/|cos x| ≥ 1 and |cosec x| = 1/|sin x| ≥ 1.
• As x runs through the four quadrants from 0 to 2π, sin x rises from 0 to 1, falls from 1 to 0, falls from 0 to −1, then rises from −1 back to 0. cos x falls from 1 to 0, falls from 0 to −1, rises from −1 to 0, then rises from 0 to 1.
• tan x increases throughout each quadrant: from 0 towards ∞ in the first, from −∞ to 0 in the second, from 0 towards ∞ in the third and from −∞ to 0 in the fourth. Here ∞ only describes a value that grows without bound as x nears an odd multiple of π/2.
• sin, cos, cosec and sec repeat their values after every interval of 2π; tan and cot repeat after every interval of π, because tan (π + x) = tan x.
• The graph of sin x is a smooth wave between −1 and 1 crossing the x-axis at every nπ; the graph of cos x is the same wave shifted, with its peaks at 2nπ. The graph of tan x is a chain of rising branches, each between two vertical lines x = (2n + 1)π/2 that it approaches but never meets.
• Periodicity reduces any angle: cos (−1020°) = cos (−1020° + 3 × 360°) = cos 60° = 1/2, and tan (25π/4) = tan (6π + π/4) = tan (π/4) = 1.

Teaching this idea? Put the simulation on your own page or LMS with this code.