Graphs of the Other Trigonometric Functions

Study Sheet

Graphs of the Other Trigonometric Functions

Tangent, cotangent, secant, and cosecant: periods, asymptotes, and transformations

The Graph of y=tan⁡xy=\tan x

Concept
Tangent: key features
θ°adjopphyp

Since tan⁡x=sin⁡xcos⁡x\tan x=\dfrac{\sin x}{\cos x}, the graph has a vertical asymptote wherever cos⁡x=0\cos x=0.

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  • Period: π\pi (the pattern repeats every π\pi units).
  • Vertical asymptotes: x=π2+nπx=\dfrac{\pi}{2}+n\pi for any integer nn (i.e. …,−π2,π2,3π2,…\ldots,-\tfrac{\pi}{2},\tfrac{\pi}{2},\tfrac{3\pi}{2},\ldots).
  • xx-intercepts: x=nπx=n\pi (where sin⁡x=0\sin x=0).
  • Key points on the middle branch: (−π4,−1)\left(-\tfrac{\pi}{4},-1\right), (0,0)(0,0), (π4,1)\left(\tfrac{\pi}{4},1\right).
  • Each branch increases from −∞-\infty to +∞+\infty between consecutive asymptotes.

Opposite, adjacent and hypotenuse are named from the angle.

Example
Example: asymptotes of y=tan⁡xy=\tan x near the origin

The two asymptotes closest to the origin come from cos⁡x=0\cos x=0, which happens at x=±π2x=\pm\dfrac{\pi}{2}. The branch through the origin lives on the interval (−π2,π2)\left(-\dfrac{\pi}{2},\dfrac{\pi}{2}\right) and passes through (0,0)(0,0) with tan⁡ ⁣(π4)=1\tan\!\left(\tfrac{\pi}{4}\right)=1.

Tip

Tip: Tangent asymptotes sit where cosine is zero. Its zeros sit where sine is zero. Sketch the asymptotes first, then draw one increasing branch between each pair.

The Graph of y=cot⁡xy=\cot x

Concept
Cotangent: key features
θ°adjopphyp

Since cot⁡x=cos⁡xsin⁡x\cot x=\dfrac{\cos x}{\sin x}, the graph has a vertical asymptote wherever sin⁡x=0\sin x=0.

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  • Period: π\pi.
  • Vertical asymptotes: x=nπx=n\pi (i.e. …,−π,0,π,2π,…\ldots,-\pi,0,\pi,2\pi,\ldots).
  • xx-intercepts: x=π2+nπx=\dfrac{\pi}{2}+n\pi (where cos⁡x=0\cos x=0).
  • Key points on a branch: (π4,1)\left(\tfrac{\pi}{4},1\right), (π2,0)\left(\tfrac{\pi}{2},0\right), (3π4,−1)\left(\tfrac{3\pi}{4},-1\right).
  • Each branch decreases from +∞+\infty to −∞-\infty between consecutive asymptotes.

Opposite, adjacent and hypotenuse are named from the angle.

Example
Example: asymptotes and intercepts of y=cot⁡xy=\cot x

Set sin⁡x=0\sin x=0: asymptotes at x=0, ±π, ±2π,…x=0,\ \pm\pi,\ \pm2\pi,\ldots. Between x=0x=0 and x=πx=\pi the branch falls from +∞+\infty to −∞-\infty, crossing the axis at its midpoint x=π2x=\dfrac{\pi}{2}.

Tip

Tip: Cotangent is the “mirror image” behavior of tangent: same period π\pi, but its asymptotes are at multiples of π\pi and each branch decreases.

The Graphs of y=sec⁡xy=\sec x and y=csc⁡xy=\csc x

Concept
Secant and cosecant as reciprocals
xy

Because sec⁡x=1cos⁡x\sec x=\dfrac{1}{\cos x} and csc⁡x=1sin⁡x\csc x=\dfrac{1}{\sin x}, you can sketch each one from its “parent” wave:

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  • An asymptote appears wherever the parent (cosine or sine) equals 00.
  • Where the parent has a maximum of 11, the reciprocal has a minimum of 11; where the parent has a minimum of −1-1, the reciprocal has a maximum of −1-1.
  • Period of each: 2π2\pi.   Range of each: (−∞,−1]∪[1,∞)(-\infty,-1]\cup[1,\infty) (no values strictly between −1-1 and 11).
  • Secant asymptotes: x=π2+nπx=\dfrac{\pi}{2}+n\pi (where cos⁡x=0\cos x=0).
  • Cosecant asymptotes: x=nπx=n\pi (where sin⁡x=0\sin x=0).

Slope is rise over run.

Example
Example: reading the secant graph

At x=0x=0, cos⁡0=1\cos 0=1, so sec⁡0=11=1\sec 0=\dfrac{1}{1}=1: the upward U touches its minimum (0,1)(0,1). At x=πx=\pi, cos⁡π=−1\cos\pi=-1, so sec⁡π=−1\sec\pi=-1: the downward U touches its maximum (π,−1)(\pi,-1). The asymptotes at x=±π2x=\pm\dfrac{\pi}{2} separate these U-shapes.

Tip

Tip: Lightly sketch the cosine (for sec⁡\sec) or the sine (for csc⁡\csc) first. Draw asymptotes through its zeros, then draw a U touching each hump. There is never any part of the graph between y=−1y=-1 and y=1y=1.

Transformations: $y=a f

(bx-c)+d$

Amplitude is the height; period is one full cycle.

Concept
Period, phase shift, and asymptotes

For b>0b>0, write the inside as b(x−cb)b\left(x-\dfrac{c}{b}\right). Then:

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  • Phase (horizontal) shift: cb\dfrac{c}{b};   vertical shift: dd;   ∣a∣|a| stretches vertically (it does not change the period).
  • Period: tangent & cotangent use π∣b∣\dfrac{\pi}{|b|};   secant & cosecant use 2π∣b∣\dfrac{2\pi}{|b|}.
  • Asymptotes come from the inside expression:
Example
Example: y=3tan⁡ ⁣(2x−π2)y=3\tan\!\left(2x-\dfrac{\pi}{2}\right)

Rewrite: 2x−π2=2 ⁣(x−π4)2x-\dfrac{\pi}{2}=2\!\left(x-\dfrac{\pi}{4}\right), so b=2b=2, phase shift =π4=\dfrac{\pi}{4} right, a=3a=3.

Period=π∣b∣=π2.\text{Period}=\frac{\pi}{|b|}=\frac{\pi}{2}.

Asymptotes: 2x−π2=π2+nπ  ⇒  2x=π+nπ  ⇒  x=π2+nπ2.2x-\dfrac{\pi}{2}=\dfrac{\pi}{2}+n\pi \;\Rightarrow\; 2x=\pi+n\pi \;\Rightarrow\; x=\dfrac{\pi}{2}+\dfrac{n\pi}{2}. The factor a=3a=3 makes the branches steeper but leaves the period unchanged.

Example
Example: y=csc⁡ ⁣(x2)y=\csc\!\left(\dfrac{x}{2}\right)

Here b=12b=\dfrac{1}{2}, so the period is 2π∣b∣=2π1/2=4π\dfrac{2\pi}{|b|}=\dfrac{2\pi}{1/2}=4\pi. Asymptotes come from x2=nπ⇒x=2nπ\dfrac{x}{2}=n\pi \Rightarrow x=2n\pi (that is, x=0,±2π,±4π,…x=0,\pm2\pi,\pm4\pi,\ldots).

Tip

Tip: To find asymptotes of a transformed graph, set the inside of the function equal to the asymptote condition for the parent (π2+nπ\tfrac{\pi}{2}+n\pi for tan⁡/sec⁡\tan/\sec, or nπn\pi for cot⁡/csc⁡\cot/\csc) and solve for xx. The domain is then “all real numbers except those xx-values.”

Going Deeper: Advanced Graph Ideas

Concept
Full analysis of y=atan⁡(bx−c)+dy=a\tan(bx-c)+d
xy

Treat the whole function as one transformed tangent. With b>0b>0, rewrite the inside as b ⁣(x−cb)b\!\left(x-\dfrac{c}{b}\right).

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  • Period: π∣b∣\dfrac{\pi}{|b|};   phase shift: cb\dfrac{c}{b};   vertical shift: dd.
  • Asymptote family: solve bx−c=π2+nπbx-c=\dfrac{\pi}{2}+n\pi, giving x=cb+π2b+nπbx=\dfrac{c}{b}+\dfrac{\pi}{2b}+\dfrac{n\pi}{b} for every integer nn. Consecutive asymptotes are exactly one period apart.
  • Range: all real numbers, (−∞,∞)(-\infty,\infty) --- the vertical stretch ∣a∣|a| and shift dd never bound a tangent.
  • Nearest “intercept” (center) point: the branch crosses its centerline y=dy=d where bx−c=nπbx-c=n\pi, i.e. x=cb+nπbx=\dfrac{c}{b}+\dfrac{n\pi}{b}. Between two asymptotes, this midpoint is the point of symmetry.

Slope is rise over run.

Concept
Full analysis of y=asec⁡(bx−c)+dy=a\sec(bx-c)+d

Sketch the guide wave y=acos⁡(bx−c)+dy=a\cos(bx-c)+d first; the secant hugs its humps.

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  • Period: 2π∣b∣\dfrac{2\pi}{|b|};   phase shift: cb\dfrac{c}{b};   vertical shift: dd.
  • Asymptote family: solve bx−c=π2+nπbx-c=\dfrac{\pi}{2}+n\pi, giving x=cb+π2b+nπbx=\dfrac{c}{b}+\dfrac{\pi}{2b}+\dfrac{n\pi}{b} (wherever the guide cosine is zero).
  • Range: the guide's minima sit at y=d−∣a∣y=d-|a| and maxima at y=d+∣a∣y=d+|a|, so the range is (−∞, d−∣a∣]∪[d+∣a∣, ∞)(-\infty,\,d-|a|]\cup[d+|a|,\,\infty). Nothing lands in the open gap (d−∣a∣, d+∣a∣)\bigl(d-|a|,\,d+|a|\bigr).
  • No xx-intercepts unless the gap contains 00, i.e. unless ∣a∣>∣d∣|a|>|d|; otherwise the graph never crosses the xx-axis.
Example
Worked example: full analysis of y=2sec⁡ ⁣(x2+π4)−1y=2\sec\!\left(\dfrac{x}{2}+\dfrac{\pi}{4}\right)-1

First force the “bx−cbx-c” form: x2+π4=12 ⁣(x+π2)\dfrac{x}{2}+\dfrac{\pi}{4}=\dfrac{1}{2}\!\left(x+\dfrac{\pi}{2}\right), so b=12b=\dfrac{1}{2}, phase shift =−π2=-\dfrac{\pi}{2} (left), a=2a=2, d=−1d=-1.

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  • Period: 2π∣b∣=2π1/2=4π\dfrac{2\pi}{|b|}=\dfrac{2\pi}{1/2}=4\pi.
  • Asymptotes: x2+π4=π2+nπ⇒x2=π4+nπ⇒x=π2+2nπ.\dfrac{x}{2}+\dfrac{\pi}{4}=\dfrac{\pi}{2}+n\pi \Rightarrow \dfrac{x}{2}=\dfrac{\pi}{4}+n\pi \Rightarrow x=\dfrac{\pi}{2}+2n\pi.
  • Range: d±∣a∣=−1±2d\pm|a|=-1\pm2, so minima at y=1y=1, maxima at y=−3y=-3: range (−∞,−3]∪[1,∞)(-\infty,-3]\cup[1,\infty).
  • Intercepts: since ∣a∣=2>∣d∣=1|a|=2>|d|=1, the gap (−3,1)(-3,1) contains 00, so the graph does cross the xx-axis --- solve 2sec⁡(⋯ )=12\sec(\cdots)=1, i.e. sec⁡(⋯ )=12\sec(\cdots)=\tfrac{1}{2}, which is impossible (∣sec⁡∣≥1|\sec|\ge1). Re-checking: sec⁡u=12\sec u=\tfrac12 has no solution, so despite the gap straddling 00, the reciprocal can never take that value. No xx-intercepts.

This is the subtle point: an xx-intercept needs 00 inside the range and a genuine secant value there; here 0∈(−3,1)0\in(-3,1) but the required sec⁡u=12\sec u=\tfrac12 is unreachable.

Example
Worked example: writing an equation from a described graph

A tangent-type curve increases left to right, has consecutive vertical asymptotes at x=1x=1 and x=5x=5, passes through its center point (3,4)(3,4), and reaches y=7y=7 one quarter-period to the right of that center. Build y=atan⁡(bx−c)+dy=a\tan(bx-c)+d.

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  • Period =5−1=4=5-1=4, and for tangent period =πb=\dfrac{\pi}{b}, so b=π4b=\dfrac{\pi}{4}.
  • Center (3,4)(3,4): the point of symmetry sits at the midpoint of the asymptotes (x=3x=3, check) and gives the vertical shift d=4d=4.
  • Phase: center at x=3x=3 means bx−c=0bx-c=0 there: π4(3)−c=0⇒c=3π4\dfrac{\pi}{4}(3)-c=0 \Rightarrow c=\dfrac{3\pi}{4}.
  • Amplitude factor: a quarter-period right of center is x=4x=4, where tan⁡\tan of the inside equals tan⁡π4=1\tan\dfrac{\pi}{4}=1. Then y=a(1)+4=7⇒a=3y=a(1)+4=7 \Rightarrow a=3.
y=3tan⁡ ⁣(π4x−3π4)+4.y=3\tan\!\left(\frac{\pi}{4}x-\frac{3\pi}{4}\right)+4.
Example
Worked example: where two reciprocal graphs meet

Find where y=sec⁡xy=\sec x meets y=csc⁡xy=\csc x on [0,2π)[0,2\pi). Set sec⁡x=csc⁡x\sec x=\csc x, i.e. 1cos⁡x=1sin⁡x\dfrac{1}{\cos x}=\dfrac{1}{\sin x}. Cross-multiplying (both denominators nonzero) gives sin⁡x=cos⁡x\sin x=\cos x, so tan⁡x=1\tan x=1.

x=π4orx=5π4.x=\frac{\pi}{4}\quad\text{or}\quad x=\frac{5\pi}{4}.

At x=π4x=\dfrac{\pi}{4}: sec⁡x=2\sec x=\sqrt2, so they meet at (π4,2)\left(\dfrac{\pi}{4},\sqrt2\right). At x=5π4x=\dfrac{5\pi}{4}: sec⁡x=−2\sec x=-\sqrt2, meeting at (5π4,−2)\left(\dfrac{5\pi}{4},-\sqrt2\right). Always discard any candidate where either cos⁡x=0\cos x=0 or sin⁡x=0\sin x=0 (an asymptote of one of the graphs).

Concept
Transformations that produce identical graphs

Different-looking formulas can trace the exact same curve. Watch for these coincidences:

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  • Cofunction shift: sec⁡ ⁣(x−π2)=csc⁡x\sec\!\left(x-\dfrac{\pi}{2}\right)=\csc x and cot⁡ ⁣(π2−x)=tan⁡x\cot\!\left(\dfrac{\pi}{2}-x\right)=\tan x; a phase shift can turn one reciprocal into another.
  • Half-period shift for tangent/cotangent: because the period is π\pi, tan⁡(x+π)=tan⁡x\tan(x+\pi)=\tan x; adding a full period changes nothing.
  • Reflection vs. shift: −tan⁡x=tan⁡(−x)-\tan x=\tan(-x), so a sign on aa can be absorbed into a horizontal reflection.
  • Sign of aa on secant: −sec⁡x=sec⁡(x−π)-\sec x=\sec(x-\pi) shifts the U's rather than needing a reflection.

So y=csc⁡xy=\csc x and y=sec⁡ ⁣(x−π2)y=\sec\!\left(x-\dfrac{\pi}{2}\right) are the same graph, even though one is built from sine and the other from cosine.

Example
Worked example: range of a shifted cosecant, and a combined domain

(a) For y=−3csc⁡ ⁣(2x)+5y=-3\csc\!\left(2x\right)+5: the guide is y=−3sin⁡(2x)+5y=-3\sin(2x)+5 with extremes 5±35\pm3, so minima 22 and maxima 88. The cosecant fills outside the humps: range (−∞,2]∪[8,∞)(-\infty,2]\cup[8,\infty).

(b) Domain of a combined expression f(x)=tan⁡x+csc⁡xf(x)=\tan x+\csc x. Exclude everything either piece forbids: tan⁡x\tan x dies where cos⁡x=0\cos x=0 (x=π2+nπx=\dfrac{\pi}{2}+n\pi) and csc⁡x\csc x dies where sin⁡x=0\sin x=0 (x=nπx=n\pi). Union of the two forbidden sets:

x≠nπ2(all integer multiples of π2).x\neq \frac{n\pi}{2}\quad(\text{all integer multiples of }\tfrac{\pi}{2}).

So the domain is all reals except x=nπ2x=\dfrac{n\pi}{2}.

Tip

Tip: For a range question on a shifted secant/cosecant, only dd and ∣a∣|a| matter --- the answer is always (−∞, d−∣a∣]∪[d+∣a∣, ∞)(-\infty,\,d-|a|]\cup[d+|a|,\,\infty). For a domain question on a sum of these functions, take the union of each piece's forbidden xx-values. For an intersection question, set the two equal, reduce to a single trig equation, then throw out any solution that lands on an asymptote of either curve.

Formulas, Proofs & Tips

Tip
Tangent, cotangent, secant, cosecant graphs
tan⁡x=sin⁡xcos⁡x (period π),sec⁡x=1cos⁡x\tan x=\frac{\sin x}{\cos x}\ (\text{period }\pi),\qquad \sec x=\frac{1}{\cos x}

What it means. These have asymptotes wherever their denominator is zero.

Example. tan⁡π4=1\tan\tfrac\pi4=1, and tan⁡x\tan x repeats every π\pi.

Why it works. tan⁡\tan blows up where cos⁡x=0\cos x=0, i.e. x=π2+kπx=\tfrac{\pi}{2}+k\pi. Its period is π\pi, not 2π2\pi, because sin⁡\sin and cos⁡\cos both change sign after half a turn and the signs cancel in the quotient.

Tip. sec⁡\sec has an asymptote wherever cos⁡\cos crosses zero, and touches ±1\pm1 wherever cos⁡\cos peaks.