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
xy−3π/2−π−π/2π/2π
Asymptotes where cos⁡x=0\cos x=0; each branch rises through a zero of sin⁡x\sin x.

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.

  • 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.

On the unit circle, tan⁡θ=yx\tan\theta=\dfrac{y}{x} is the slope of the radius to the point. As θ\theta climbs from 00 toward π2\tfrac{\pi}{2} the radius tilts up toward vertical, so its slope grows without bound, which is the asymptote. Just past π2\tfrac{\pi}{2} the radius tilts slightly left and its slope is a huge negative number, so the next branch begins at −∞-\infty. After half a turn the radius points the opposite way along the same line, with the same slope: period π\pi.

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.

Side note
Where it goes wrong

Giving tangent a period of 2π2\pi. It repeats every π\pi, because tan⁡(x+π)=−sin⁡x−cos⁡x=tan⁡x\tan(x + \pi) = \tfrac{-\sin x}{-\cos x} = \tan x.

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.

Try it
Quick check

On [0,2π][0,2\pi], list the vertical asymptotes and the xx-intercepts of y=tan⁡xy=\tan x, and give tan⁡5π4\tan\dfrac{5\pi}{4}.

Show the answer

Answer: asymptotes x=π2, 3π2x=\dfrac{\pi}{2},\ \dfrac{3\pi}{2}; intercepts x=0, π, 2πx=0,\ \pi,\ 2\pi; tan⁡5π4=1\tan\dfrac{5\pi}{4}=1. Asymptotes are the zeros of cos⁡x\cos x and intercepts the zeros of sin⁡x\sin x. And 5π4=π4+π\dfrac{5\pi}{4}=\dfrac{\pi}{4}+\pi, one period past π4\dfrac{\pi}{4}, so the value repeats: 11.

The Graph of y=cot⁡xy=\cot x

Concept
Cotangent: key features
xy−3−2−112π/2π3π/22π
Asymptotes at multiples of π\pi (including the yy-axis); each branch falls.

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.

  • 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.

Cotangent is xy\dfrac{x}{y} for the unit-circle point, the reciprocal of the slope. Where tan⁡x\tan x is 00 its reciprocal blows up, and where tan⁡x\tan x blows up its reciprocal is 00; so the asymptotes and zeros of tangent swap roles. Each branch falls because the reciprocal of an increasing positive quantity decreases.

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.

Try it
Quick check

On (0,π)(0,\pi), where is cot⁡x=−1\cot x=-1? What happens to cot⁡x\cot x as x→πx\to\pi from the left?

Show the answer

Answer: x=3π4x=\dfrac{3\pi}{4}; cot⁡x→−∞\cot x\to-\infty. cot⁡3π4=−2/22/2=−1\cot\dfrac{3\pi}{4}=\dfrac{-\sqrt2/2}{\sqrt2/2}=-1. Near π\pi from the left, cos⁡x≈−1\cos x\approx-1 while sin⁡x\sin x is a tiny positive number, so cos⁡xsin⁡x\dfrac{\cos x}{\sin x} is a huge negative number: the branch falls to the asymptote at x=πx=\pi.

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

Concept
Secant and cosecant as reciprocals
xy−3−2−112π/2π3π/22π
y=csc⁡xy=\csc x (solid) over its guide y=sin⁡xy=\sin x (dashed): the U's touch the humps and blow up at the zeros.

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:

  • 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).

Reciprocals turn small into large and 11 into 11. Where the guide wave equals ±1\pm1 its reciprocal equals it too, so the two graphs touch there. As the guide shrinks toward 00, its reciprocal grows without bound, which is the asymptote. And because the guide never exceeds 11 in size, its reciprocal is never smaller than 11 in size: the band −1<y<1-1<y<1 stays empty.

Example
Example: reading the secant graph
xy−3−2−112−π/2π/2π
The upward U touches (0,1)(0,1), the downward U touches (π,−1)(\pi,-1).

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.

Try it
Quick check

For y=csc⁡xy=\csc x: state the range, list the asymptotes on [0,2π][0,2\pi], and find the value at x=3π2x=\dfrac{3\pi}{2}.

Show the answer

Answer: range (−∞,−1]∪[1,∞)(-\infty,-1]\cup[1,\infty); asymptotes x=0, π, 2πx=0,\ \pi,\ 2\pi; csc⁡3π2=−1\csc\dfrac{3\pi}{2}=-1. The asymptotes are the zeros of sin⁡x\sin x. At 3π2\dfrac{3\pi}{2} the sine has its minimum −1-1, so the cosecant has the top of a downward U there, also −1-1.

Transformations: y=a f(bx−c)+dy=a\,f(bx-c)+d

Here ff is any of tan⁡\tan, cot⁡\cot, sec⁡\sec, csc⁡\csc. The four numbers do the same jobs they do for sine and cosine; only the period formula and the asymptotes are new.

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:

  • 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: set bx−c=π2+nπbx-c=\tfrac{\pi}{2}+n\pi for tan⁡\tan and sec⁡\sec, or bx−c=nπbx-c=n\pi for cot⁡\cot and csc⁡\csc, and solve for xx.

The parent tangent repeats every π\pi, so tan⁡(bx)\tan(bx) repeats once bxbx has advanced by π\pi, which takes πb\dfrac{\pi}{b} units of xx; the same reasoning with 2π2\pi gives the secant and cosecant periods. The asymptotes are wherever the inside lands on a forbidden value of the parent, which is why they are found by solving an equation about bx−cbx-c rather than about xx.

Example
Example: y=3tan⁡ ⁣(2x−π2)y=3\tan\!\left(2x-\dfrac{\pi}{2}\right)
xy−6−33π/2π3π/2
Asymptotes π2\tfrac{\pi}{2} apart (at 0,π2,π,…0,\tfrac{\pi}{2},\pi,\dots); each branch is centred a quarter of the way along, at π4+nπ2\tfrac{\pi}{4}+\tfrac{n\pi}{2}.

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)
xy−3−2−112π2π3π4π
Period 4π4\pi: asymptotes at 0, 2π, 4π0,\ 2\pi,\ 4\pi; the guide is sin⁡x2\sin\tfrac{x}{2} (dashed).

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.”

Try it
Quick check

For y=tan⁡x3y=\tan\dfrac{x}{3}, find the period and all vertical asymptotes.

Show the answer

Answer: period 3π3\pi; asymptotes x=3π2+3nπx=\dfrac{3\pi}{2}+3n\pi. Period =π1/3=3π=\dfrac{\pi}{1/3}=3\pi. Asymptotes: x3=π2+nπ\dfrac{x}{3}=\dfrac{\pi}{2}+n\pi, so x=3π2+3nπx=\dfrac{3\pi}{2}+3n\pi (for example ±3π2\pm\dfrac{3\pi}{2}, 9π2\dfrac{9\pi}{2}); consecutive ones are one period apart, as they must be.

Going Deeper: Advanced Graph Ideas

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

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).

  • 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.
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.

  • 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).
  • xx-intercepts exist exactly when 00 is in the range, i.e. when ∣d∣≥∣a∣|d|\ge|a|. If ∣a∣>∣d∣|a|>|d| the gap contains 00 and the graph never touches 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
xy−6−4−224−ππ2π3π
Asymptotes at x=π2+2nπx=\tfrac{\pi}{2}+2n\pi; nothing lies between y=−3y=-3 and y=1y=1, so the graph misses the xx-axis.

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.

  • 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: y=0y=0 would need 2sec⁡(⋯ )−1=02\sec(\cdots)-1=0, i.e. sec⁡(⋯ )=12\sec(\cdots)=\tfrac12, impossible because ∣sec⁡u∣≥1|\sec u|\ge1. Equivalently, 00 lies in the gap (−3,1)(-3,1), so it is not in the range. No xx-intercepts.

In general a transformed secant crosses the xx-axis exactly when 00 is in its range, which happens when ∣d∣≥∣a∣|d|\ge|a|; here ∣a∣=2>∣d∣=1|a|=2>|d|=1.

Example
Worked example: writing an equation from a described graph
xy123456−448(3, 4)(4, 7)
One branch between x=1x=1 and x=5x=5, centred at (3,4)(3,4).

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.

  • 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
xy−4−22π/2π3π/2
sec⁡x\sec x (solid) and csc⁡x\csc x (dashed) cross at (π4,2)\left(\tfrac{\pi}{4},\sqrt2\right) and (5π4,−2)\left(\tfrac{5\pi}{4},-\sqrt2\right).

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:

  • 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.

Try it
Quick check

Find the range of y=4sec⁡(3x)−1y=4\sec(3x)-1. Does its graph have any xx-intercepts?

Show the answer

Answer: range (−∞,−5]∪[3,∞)(-\infty,-5]\cup[3,\infty); no xx-intercepts. The guide 4cos⁡(3x)−14\cos(3x)-1 runs between −1−4=−5-1-4=-5 and −1+4=3-1+4=3, and the secant fills everything outside that band. Since 00 is inside the gap (−5,3)(-5,3), the graph never reaches y=0y=0 (it would need sec⁡(3x)=14\sec(3x)=\tfrac14).

Why It Works

Tangent is the slope of the radius to the unit-circle point: tan⁡θ=sin⁡θcos⁡θ=yx\tan\theta = \tfrac{\sin\theta}{\cos\theta} = \tfrac yx. Rotating by π\pi sends (x,y)(x, y) to (−x,−y)(-x, -y), the same slope — hence period π\pi. The asymptotes are where the radius is vertical (x=0x = 0) and the slope is infinite. Secant and cosecant inherit their asymptotes from the zeros of cosine and sine, and can never lie strictly between −1-1 and 11 because a reciprocal of a number in [−1,1][-1, 1] has absolute value at least 11.

Tip
Reading a proof

Do not read a proof the way you read a rule. Cover the last line, ask what the previous line forces, and only then uncover it. If a step feels like magic, that is the step to work by hand with a concrete number.

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.