Secant: build a reciprocal graph from cosine

Explain it like I am five

Picture laying tracing paper over the cosine wave and redrawing it with one rule: keep each input x and replace each height h with 1 ÷ h, its reciprocal. The new drawing is secant, sec x = $1 over cos x$ .

Walk along it. At x = 0 cosine is 1, so secant is 1: the bottom of an upward U. As x moves toward $π over 2$ (about 1.57), cosine shrinks through 0.5, 0.1 and 0.01, and secant grows through 2, 10 and 100. At x = $π over 2$ cosine is 0, so secant has no value: draw a dashed vertical asymptote there. Past it cosine is negative, so secant is negative; at x = π cosine is −1, so secant is −1, the top of an upside-down U.

Asymptotes stand wherever cos x = 0, at x = $π over 2$ + nπ. The branches never enter the band between −1 and 1, the pattern repeats every 2π, and the graph matches its mirror image across the y-axis because secant is even.

In plain words

Think of the same height-changing machine with a different starting drawing. This time you feed it cosine heights. It leaves each horizontal position alone and replaces the height with one divided by that height. At the center, cosine has height one, so secant also has height one. Half a turn farther along, cosine has height negative one, so secant has height negative one. Zero heights still cannot pass through the machine. The finished picture has a curved positive branch centered at zero and a curved negative branch centered at π. The pattern repeats after a full turn.

The horizontal axis uses radians; the dashed cosine guide locates secant's asymptotes and its shared heights of 1 and −1.

Reminder

Coordinates. A point lists input then height: (π, −1) means input π and output −1.
Radians. Graph inputs use radians: π = 180° and $3 π over 2$ = 270°.
Unit-circle cosine. Cosine is the unit-circle horizontal coordinate: cos( $π over 3$ ) = $1 over 2$ .
Reciprocal graph. Keep the input and take one divided by a nonzero height: ( $π over 3$ , $1 over 2$ ) becomes ( $π over 3$ , 2).
Fraction division. Dividing by a fraction multiplies by its reciprocal: 1 ÷ $1 over 2$ = 1 × $2 over 1$ = 2.
Negative division. One divided by a negative stays negative: 1 ÷ (− $1 over 2$ ) = −2.
Rationalizing a denominator. Multiply top and bottom by the same root: $2 over square root of 2$ = $2 square root of 2 over 2$ = $square root of 2$ .
Zero denominator. Exclude an input that makes the bottom zero: cos( $π over 2$ ) = 0 makes sec( $π over 2$ ) undefined.
Integer multiples. In $π over 2$ + nπ, n may be any integer: n = −1 gives − $π over 2$ , and n = 1 gives $3 π over 2$ .
Intervals and union. Parentheses exclude boundaries, and ∪ joins permitted pieces: (− $π over 2$ , $π over 2$ ) excludes both asymptote inputs.
Range. The reciprocal range includes its edge heights: y ≤ −1 or y ≥ 1 allows −1 and 1 but excludes $1 over 2$ .
Even symmetry. An even function keeps its height at opposite inputs: sec(− $π over 3$ ) = sec( $π over 3$ ) = 2.
Period. A 2π period repeats a height: sec(−π + 2π) = sec π = −1.
Vertical asymptote. A shrinking nonzero denominator produces unbounded nearby heights: cosine approaching zero makes secant's magnitude grow.

Why it works. Secant comes from sec x = . Cosine is zero at x = + nπ, so those inputs make the denominator zero and create the asymptotes. Its heights stay between −1 and 1, giving reciprocal heights at or beyond ±1. Cosine is even, so opposite inputs have equal cosine values and equal reciprocals. Cosine repeats every 2π, and its reciprocal repeats with it. A shift by π changes the sign instead of repeating a nonzero height.

RuleRule: sec x = , with x ≠ + nπ for every integer n. Its vertical asymptotes are x = + nπ, its range is (−∞, −1] ∪ [1, ∞), its period is 2π, and sec(−x) = sec x.

The same idea, five ways

Say it

Say: Secant of x is one divided by cosine of x.

Write it

Write: A secant graph takes reciprocal cosine heights at unchanged inputs.

In math

sec x = $1 over cos x$
Domain: {x | x is real and x ≠ $π over 2$ + nπ, for every integer n}
Range: y ≤ −1 or y ≥ 1
Range: (−∞, −1] ∪ [1, ∞)
sec(x + 2π) = sec x
sec(−x) = sec x
Graph words: reciprocal cosine branches with symmetry about the y-axis

Like: using a different set of height measurements in the same reciprocal machine.

See it

The horizontal axis uses radians; the dashed cosine guide locates secant's asymptotes and its shared heights of 1 and −1.

The same idea, other ways

As a cosine-guided picture

The cosine wave begins at (0, 1), so secant's positive vertex begins there too. The zeros at − $π over 2$ and $π over 2$ frame that branch. Cosine's bottom at (π, −1) locates the next negative vertex.

The horizontal axis uses radians; the dashed cosine guide locates secant's asymptotes and its shared heights of 1 and −1.

With a changed measurement

Cosine height $1 over 2$ becomes secant height 2 because $1 over 2$ × 2 = 1. Cosine height − $1 over 2$ becomes −2 because (− $1 over 2$ ) × (−2) = 1. Neither input nor sign changes.

cos x = → sec x = 2

cos x = − → sec x = −2

Reciprocal heights keep their sign and multiply with their guide heights to give 1.

As a mirror check

Cosine gives the same height at x and −x. Taking the reciprocal of equal heights gives equal answers. Thus the secant points ( $π over 3$ , 2) and (− $π over 3$ , 2) match across the y-axis.

The horizontal axis uses radians. Cosine reaches 1 at zero and zero at the boundary lines; the positive secant branch stays at height 1 or above.

.1The positive secant branch centered at 0

Picture a bowl directly above the origin. Its lowest point sits at height one, and the two sides match across the y-axis. This comes from the cosine wave, whose highest point is at input zero. As you walk left or right from zero toward the nearest cosine zeros, the positive cosine height shrinks. Its reciprocal therefore rises. The boundary inputs are negative π over two and positive π over two, and neither is part of secant's branch.

Rule: on − $π over 2$ < x < $π over 2$ , sec x ≥ 1.
Rule: the vertex is (0, 1), shared with cosine.
Rule: x = − $π over 2$ and x = $π over 2$ are the bounding asymptotes; opposite inputs have equal heights.

The horizontal axis uses radians. Cosine reaches 1 at zero and zero at the boundary lines; the positive secant branch stays at height 1 or above.

Reminder

Reciprocal signs. 1 ÷ $1 over 2$ = 2, and 1 ÷ (− $1 over 2$ ) = −2; the sign stays the same.

The same idea, five ways

Say it

Say: Between negative pi over two and pi over two, secant is one or greater.

Write it

Write: The positive branch has its lowest point at input zero and is symmetric about the y-axis.

In math

− $π over 2$ < x < $π over 2$
x ∈ (− $π over 2$ , $π over 2$ )
sec x ≥ 1
Vertex: (0, 1)
sec(−x) = sec x
Graph words: upward-opening curved branch centered at 0

Like: a bowl centered above the origin, with its bottom one unit above the road.

See it

The horizontal axis uses radians. Cosine reaches 1 at zero and zero at the boundary lines; the positive secant branch stays at height 1 or above.

Worked exampleThree points on secant's central branch

You need the central turning point and a matching point on each side. Sketch the secant branch on − $π over 2$ < x < $π over 2$ .

The horizontal axis uses radians. Cosine reaches 1 at zero and zero at the boundary lines; the positive secant branch stays at height 1 or above.

Draw the two boundary lines and check cos(− $π over 2$ ) = cos( $π over 2$ ) = 0.These two zero-denominator inputs are excluded and locate the asymptotes.
Mark (0, 1).The guide height is cos 0 = 1, whose reciprocal is 1.
Mark ( $π over 3$ , 2).Cosine at 60° is $1 over 2$ , and 1 ÷ $1 over 2$ = 2.
Mark (− $π over 3$ , 2) and connect both sides through (0, 1), bending upward toward the boundary lines.Secant is even, so the opposite input has the same height; both sides grow as the positive cosine denominator shrinks.

Answer

Interval: (− $π over 2$ , $π over 2$ ).
Vertex: (0, 1).
Supporting points: (− $π over 3$ , 2) and ( $π over 3$ , 2).
Shape: one upward-opening positive branch.

Check Independent unit-circle check: the point at −60° has horizontal coordinate , giving sec(−) = 2 directly. That agrees with the point found using even symmetry.

Counterexamples: what it is not, and the tempting wrong moves

✗ Not this: Put the positive secant vertex at (, 1).

Cosine is zero there, so secant is undefined at that input.

✓ Instead: The central positive vertex is (0, 1), where cosine equals 1.

Tips and tricks

Start at the cosine peak (0, 1), then build matching reciprocal points on both sides.

.2The negative secant branch centered at π

Picture an upside-down bowl whose middle sits below the input π. Its highest point has height negative one. The cosine guide is negative between π over two and three π over two, so all the reciprocal heights there are negative too. At π the guide reaches negative one, which keeps its value after reciprocation. To either side, cosine moves toward zero while staying negative. Secant then moves downward, farther from the x-axis, approaching the two boundary lines.

Rule: on $π over 2$ < x < $3 π over 2$ , sec x ≤ −1.
Rule: the vertex is (π, −1), shared with cosine.
Rule: x = $π over 2$ and x = $3 π over 2$ are the bounding asymptotes, and the branch opens downward.

The horizontal axis uses radians. Cosine reaches −1 at π and zero at the boundary lines; the negative secant branch stays at height −1 or below.

Reminder

Reciprocal signs. 1 ÷ $1 over 2$ = 2, and 1 ÷ (− $1 over 2$ ) = −2; the sign stays the same.

The same idea, five ways

Say it

Say: Between pi over two and three pi over two, secant is negative one or less.

Write it

Write: The negative branch has its highest point at input pi.

In math

$π over 2$ < x < $3 π over 2$
x ∈ ( $π over 2$ , $3 π over 2$ )
sec x ≤ −1
Vertex: (π, −1)
Graph words: downward-opening curved branch centered at π

Like: an upside-down bowl hanging below the input pi.

See it

The horizontal axis uses radians. Cosine reaches −1 at π and zero at the boundary lines; the negative secant branch stays at height −1 or below.

Worked exampleThree points on secant's negative branch

You need the turning point and two supporting points for the negative piece. Sketch the secant branch on $π over 2$ < x < $3 π over 2$ .

The horizontal axis uses radians. Cosine reaches −1 at π and zero at the boundary lines; the negative secant branch stays at height −1 or below.

Draw the two boundary lines and verify cos( $π over 2$ ) = cos( $3 π over 2$ ) = 0.This confirms the denominator zeros that separate the negative branch from neighboring branches.
Mark (π, −1).Cos π = −1, and $1 over −1$ = −1.
Mark ( $2 π over 3$ , −2) and ( $4 π over 3$ , −2).At 120° and 240° the unit-circle horizontal coordinates are − $1 over 2$ , whose reciprocal is −2. Rebuild them from $π over 3$ : $2 π over 3$ = π − $π over 3$ reverses the horizontal coordinate; $4 π over 3$ = − $2 π over 3$ + 2π keeps that negative cosine.
Connect the points with a downward-opening curve approaching both boundary lines.Cosine stays negative and approaches zero at both ends, so secant becomes negative with unbounded magnitude.

Answer

Interval: ( $π over 2$ , $3 π over 2$ ).
Vertex: (π, −1).
Supporting points: ( $2 π over 3$ , −2) and ( $4 π over 3$ , −2).
Shape: one downward-opening negative branch.

Check Independent symmetry and period check: reflecting (, −2) across the y-axis gives (−, −2). Adding 2π to that input gives (, −2), which matches the other supporting point.

Counterexamples: what it is not, and the tempting wrong moves

✗ Not this: Secant's negative branch can reach −.

That height would require cosine −2, beyond a unit-circle coordinate.

✓ Instead: The branch stays at or below −1, with its highest point (π, −1).

Tips and tricks

Check negative supporting points by multiplying their height by the cosine guide height.

Strategy: step by step

1. Lightly sketch y = cos x as the guide wave. Keep its horizontal inputs fixed.
2. Draw a vertical asymptote at every x-intercept of cosine. Set cos x = 0 to locate the forbidden denominator inputs, then plug them back into cosine to confirm zero.
3. Keep the points where cosine is 1 or −1. They are also points on secant and become the branch vertices.
4. Between asymptotes draw a curved U shaped branch opening away from the x-axis. The branch never enters −1 < y < 1. Use reciprocal cosine heights to place the curve; its U shape describes the bend rather than giving a different equation.
5. Repeat the two-branch pattern every 2π. Use the y-axis symmetry to check matching points at opposite inputs.

Strategy

Strategy: build secant from a cosine guide

Is the cosine height zero?

YesSet cos x = 0 to find the forbidden input, plug it back into cosine to verify zero, then draw a vertical asymptote rather than a secant point.

NoKeep the input and take one divided by the cosine height.

↓

Is cosine positive on this interval?

YesDraw a positive upward-opening branch with heights y ≥ 1.

NoDraw a negative downward-opening branch with heights y ≤ −1.

↓

Is the point reflected to the opposite input?

YesKeep its secant height unchanged, because secant is even.

NoUse the reciprocal cosine height at the original input.

1. Identify cosine zeros, because they are forbidden secant inputs.
2. Mark cosine's heights 1 and −1 as the secant vertices.
3. Take reciprocal heights at one or two additional inputs on each branch.
4. Connect the points with curves facing away from the x-axis and copy the pattern by 2π.

Worked exampleA full cosine-guided secant period

You need one complete positive branch and one complete negative branch. Sketch y = sec x on − $π over 2$ ≤ x ≤ $3 π over 2$ using cosine as the guide.

The horizontal axis uses radians; the dashed cosine guide locates secant's asymptotes and its shared heights of 1 and −1.

Read the cos x height under its input angle; all inputs in this picture are radians.

Read the sec x height under its input angle; all inputs in this picture are radians.

Draw the cosine guide using the cosine fntable visual.The starting heights appear directly below the angle inputs.
Set cos x = 0 to locate forbidden inputs: − $π over 2$ , $π over 2$ , $3 π over 2$ . Plug them back in: cos(− $π over 2$ ) = cos( $π over 2$ ) = cos( $3 π over 2$ ) = 0.These values make the secant denominator zero, so each is excluded and receives an asymptote.
Keep (0, 1) and (π, −1).Cosine has heights 1 and −1 there, and their reciprocals have the same values.
Read the columns under − $π over 3$ and $π over 3$ . Change each cosine height $1 over 2$ to secant height 2.The positive branch lies between − $π over 2$ and $π over 2$ , and these points guide its two sides.
Read the columns under $2 π over 3$ and $4 π over 3$ . Change each cosine height − $1 over 2$ to secant height −2.The negative branch lies between $π over 2$ and $3 π over 2$ , and both support points remain negative. $2 π over 3$ = π − $π over 3$ , so vertical-axis reflection reverses cosine. $4 π over 3$ = − $2 π over 3$ + 2π; reflection across the horizontal axis and a full turn both keep that cosine.
Draw an upward-opening curved branch centered at 0 and a downward-opening curved branch centered at π.Near either boundary the cosine height approaches zero with the branch's sign, so the reciprocal moves away from the x-axis without bound.

Answer

Asymptotes in the interval: x = − $π over 2$ , $π over 2$ , $3 π over 2$ .
Positive branch: − $π over 2$ < x < $π over 2$ , vertex (0, 1).
Negative branch: $π over 2$ < x < $3 π over 2$ , vertex (π, −1).
The interval length and the repeating pattern are 2π.

Check Independent symmetry check: reflecting (, 2) across the y-axis gives (−, 2), the plotted positive point. Reflecting (π, −1) gives (−π, −1); adding the period 2π returns to (π, −1). The heights also satisfy the two allowed range bands.

Ladder: from easy to exam-hard. Press Try it first on any rung to hide its steps and use them as hints.

Rung 1Rung 1: secant at the origin's input

You need the secant height at input zero. Evaluate sec 0 and locate its point.

The horizontal axis uses radians. Cosine reaches 1 at zero and zero at the boundary lines; the positive secant branch stays at height 1 or above.

Use cos 0 = 1.The unit-circle point at 0° is (1, 0), and cosine is its horizontal coordinate.
Compute sec 0 = $1 over 1$ = 1.Secant is the reciprocal of cosine.
Plot (0, 1).The unchanged input is zero and the new output is 1.

Answer

sec 0 = 1.
Point: (0, 1).

Check Multiplication check: cos 0 × sec 0 = 1 × 1 = 1. This point also lies on the y-axis, so even symmetry leaves it in place.

Rung 2Rung 2: the negative branch at and beside its vertex

You need the negative branch's central height and one neighboring height. Evaluate sec π and sec( $2 π over 3$ ).

The horizontal axis uses radians. Cosine reaches −1 at π and zero at the boundary lines; the negative secant branch stays at height −1 or below.

Use cos π = −1 and cos( $2 π over 3$ ) = − $1 over 2$ .The unit-circle horizontal coordinates at 180° and 120° give these exact cosine values.
Compute sec π = $1 over −1$ = −1.The reciprocal of −1 is −1, locating the negative vertex.
Compute sec( $2 π over 3$ ) = 1 ÷ (− $1 over 2$ ) = −2.Dividing one by a negative half gives a negative output of magnitude 2.
Plot (π, −1) and ( $2 π over 3$ , −2) below the x-axis.Both inputs lie in the cosine-negative interval ( $π over 2$ , $3 π over 2$ ), so their secant heights are negative.

Answer

sec π = −1.
sec( $2 π over 3$ ) = −2.

Check Multiplication checks: (−1) × (−1) = 1 and (−) × (−2) = 1. The second point is farther below the axis than the vertex, as required.

Rung 3Rung 3: a full period with a negative-input branch

You need two complete secant branches including the one centered at negative pi. Sketch y = sec x on − $3 π over 2$ ≤ x ≤ $π over 2$ and state the usable inputs within this interval.

The horizontal axis uses radians; the dashed cosine guide locates secant's asymptotes and its shared heights of 1 and −1.

Lightly sketch cosine, then solve cos x = 0 to locate the forbidden inputs − $3 π over 2$ , − $π over 2$ , $π over 2$ . Plug them back into cosine and verify that all three values are zero.This finds the zero-denominator inputs so they can be excluded and marked as asymptotes.
Keep the guide-wave points (−π, −1) and (0, 1).Cosine has heights −1 and 1 there, so the reciprocal heights remain −1 and 1.
Add (− $2 π over 3$ , −2) on the negative branch and (− $π over 3$ , 2) on the positive branch.The guide heights are − $1 over 2$ and $1 over 2$ , and their reciprocals are −2 and 2.
Draw a downward-opening branch centered at −π and an upward-opening branch centered at 0, each approaching its bounding asymptotes.The branches keep cosine's signs and stay outside the band −1 < y < 1.
Write the usable inputs as (− $3 π over 2$ , − $π over 2$ ) ∪ (− $π over 2$ , $π over 2$ ).Open endpoints remove the three denominator zeros while keeping all remaining inputs in the interval.

Answer

Asymptotes: x = − $3 π over 2$ , − $π over 2$ , $π over 2$ .
Vertices: (−π, −1) and (0, 1).
Usable inputs: (− $3 π over 2$ , − $π over 2$ ) ∪ (− $π over 2$ , $π over 2$ ).

Check Independent repetition check: adding 2π to (−π, −1) gives (π, −1), the previously drawn negative vertex. Even symmetry also matches (−, 2) with (, 2). The requested interval has length 2π.

Counterexamples: what it is not, and the tempting wrong moves

✗ Not this: Counterexample, wrong: put secant's asymptotes at x = 0 and x = π because those are sine zeros.

Why it fails: secant divides by cosine. Cos 0 = 1 and cos π = −1, so both denominators are nonzero and both inputs are allowed.

✓ Instead: Right: x = 0 and x = π give vertices with heights 1 and −1. Secant's asymptotes are at x = + nπ.

✗ Not this: Counterexample, wrong: reflect (, 2) to (−, −2) for secant.

Why it fails: that changes both signs and would use odd symmetry. Secant is even because cosine at opposite inputs is equal.

✓ Instead: Right: reflection across the y-axis gives (−, 2), with the same height.

✗ Not this: Counterexample: repeat a positive secant branch after π with its height unchanged.

sec 0 = 1 but sec π = −1, so π changes the sign. The same-height pattern returns after 2π.

✓ Instead: Right: use a full 2π period containing one positive branch and one negative branch, each bending away from the x-axis toward its boundary lines.

Tips and tricks

Tip: 'sec pairs with cos' identifies the guide wave and the denominator to check.
Tip: the first secant vertex is (0, 1); the next is (π, −1). These two points distinguish secant from cosecant.
Tip: opposite secant inputs have the same height. Use this y-axis symmetry to check the central branch.
Tip: touching the guide at ±1 does not make the whole curves equal. All smaller nonzero guide magnitudes produce reciprocal magnitudes greater than 1.
Tip: mark all cosine zeros before drawing any branch, then keep every branch outside −1 < y < 1.

Trap. Trap: secant uses cosine zeros, which lie at odd multiples of . Cosecant uses sine zeros at multiples of π. Check the denominator before placing any asymptote.

Keep in mind

Secant's asymptotes come from cosine's zeros, $π over 2$ + nπ, not sine's: x = 0 is a vertex of secant, not an asymptote.
Secant is even, so its graph is its own mirror image across the y-axis: sec(−1.2) = sec 1.2 ≈ 2.76.
Secant and cosecant have the same U shapes, shifted: secant's first upward vertex is at x = 0, cosecant's at x = $π over 2$ .

Memory hookSecant sits on cosine: its U shapes rest on cosine's peaks and valleys, and its walls stand where cosine crosses zero, at + nπ.

Flash cards: say the answer out loud, then flip

Which wave is the guide for sec x, and why?

cos x, because sec x = .

Where are secant's vertical asymptotes?

x = + nπ, where cos x = 0.

cos x = 0.4. What is sec x?

2.5, because 1 ÷ 0.4 = 2.5.

cos x = −. What is sec x?

− = −1.6.

Is sec x even or odd, and what does that do to its graph?

Even: the graph matches its mirror image across the y-axis.

True or false: secant's asymptotes are at x = nπ, like cosecant's.

False: secant divides by cosine, which is 0 at + nπ. At nπ, cosine is 1 or −1.