DWG NO. 4.5 — Lesson 5 of 6

Graphing Other Trig Functions

Unit 4: Trigonometric Functions and the Unit Circle · ~25 min

Objective Graph y = tan x, cot x, sec x, and csc x, identifying their periods, asymptotes, and domain restrictions.

Tangent, cotangent, secant, and cosecant are all built from sine and cosine as ratios — and wherever a denominator hits zero, the function is undefined and the graph shoots off toward infinity. That gives all four of these graphs a completely different look from the smooth, bounded waves of sine and cosine: vertical asymptotes and unbounded range.

Definitions and key features

Function	Formula	Period	Asymptotes at	Range
tan x	sin \(\frac{x}{cos}\) x	π	x = \(\frac{\pi}{2}\) + nπ	(−∞, ∞)
cot x	cos \(\frac{x}{sin}\) x	π	x = nπ	(−∞, ∞)
sec x	\(\frac{1}{cos}\) x	2π	x = \(\frac{\pi}{2}\) + nπ	(−∞,−1] ∪ [1,∞)
csc x	\(\frac{1}{sin}\) x	2π	x = nπ	(−∞,−1] ∪ [1,∞)

Notice that tan x and cot x have asymptotes exactly where cos x and sin x are zero (since those sit in the denominator), and the same is true for sec x and csc x. sec x and csc x also never take values between −1 and 1 — the reciprocal of a number with |value| ≤ 1 always has |value| ≥ 1.

y = tan x — period π, with vertical asymptotes (dashed) wherever cos x = 0

Worked Example 1 · Asymptotes of tan x

ProblemFind the vertical asymptotes of y = tan x on (−π, π).

1tan x = sin \(\frac{x}{cos}\) x is undefined where cos x = 0. On this interval, cos x = 0 at x = −\(\frac{\pi}{2}\) and x = \(\frac{\pi}{2}\).

Asymptotes at x = −\(\frac{\pi}{2}\) and x = \(\frac{\pi}{2}\)

Worked Example 2 · Evaluating sec x

ProblemFind sec \(\frac{2\pi}{3}\).

1cos \(\frac{2\pi}{3}\) = \(\frac{-1}{2}\) (Quadrant II, reference angle \(\frac{\pi}{3}\)).

2sec x = \(\frac{1}{cos x}\), so sec \(\frac{2\pi}{3}\) = \(\frac{1}{\frac{-1}{2}}\) = −2.

sec \(\frac{2\pi}{3}\) = −2

Worked Example 3 · Range check

ProblemExplain why csc x = 0.5 has no solution.

1csc x = \(\frac{1}{sin}\) x, and sin x is always between −1 and 1.

2Taking the reciprocal of any number with |value| ≤ 1 gives |value| ≥ 1, so csc x can never fall strictly between −1 and 1.

0.5 is outside the range of csc x — no solution exists

Guided practice

1.Find the vertical asymptotes of y = cot x on (0, 2π).
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cot x is undefined where sin x = 0: x = π (and the endpoints x = 0, x = 2π).
2.Find csc \(\frac{\pi}{6}\).
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sin \(\frac{\pi}{6}\) = \(\frac{1}{2}\), so csc \(\frac{\pi}{6}\) = \(\frac{1}{\frac{1}{2}}\) = 2.
3.What is the period of y = sec x, and how does it compare to the period of y = cos x?
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Period 2π — the same as cos x, since sec x is built directly from it.
4.State the range of y = sec x.
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(−∞, −1] ∪ [1, ∞) — sec x never lands strictly between −1 and 1.
5.Explain why y = tan x has period π while y = sin x has period 2π, even though tan x is built from sin x.
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tan x = sin \(\frac{x}{cos}\) x. After a half-rotation (adding π), both sin x and cos x flip sign — sin(x+π) = −sin x and cos(x+π) = −cos x — so their ratio is unchanged: tan(x+π) = \(\frac{-sin x}{-cos x}\) = tan x. The sign flips cancel, so tan x repeats twice as often as sine alone.

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