Range: the output heights each graph can reach

Explain it like I am five

Picture two glass elevators in an endless tower, each able to stop at any height, even between floors. The tangent and cotangent elevator reaches every height. The secant and cosecant elevator never stops strictly between −1 and 1, though it does stop at 1 and −1.

Why: sec x = $1 over cos x$ and csc x = $1 over sin x$ , and cosine and sine are never bigger than 1 in size, so 1 divided by either is at least 1 in size. Tangent is $\frac{\sin x}{\cos x}$ ; near x = $π over 2$ the bottom gets tiny, so the height passes any level you name.

Example: can csc x = 3? Flip it back: sin x would be $1 over 3$ , about 0.33, an allowed sine, so yes. Can csc x = −0.5? Flip: sin x would be 1 ÷ (−0.5) = −2, below −1, so no.

So the range, all output heights, of tan and cot is (−∞, ∞): every real number (∞ is infinity). For sec and csc it is (−∞, −1] ∪ [1, ∞), read 'at most −1 or at least 1'.

In plain words

Imagine the graph shining a light sideways onto a wall. Its shadow marks every height the graph ever reaches. That collection of output heights is the Range. Tangent and cotangent reach every real height, positive, zero or negative. Secant and cosecant leave a gap around zero because they are reciprocals of numbers no larger than one in size. Their heights can be 1 or more, or −1 or less. You need both the included edge heights and the missing middle. A graph window shows only part of the shadow; its top border does not put a ceiling on the function.

Inputs run horizontally in radians; outputs run vertically.

Reminder

Reciprocal signs. The reciprocal of − $1 over 2$ is −2, with the same sign.
Intervals and union. The bracket in [1, ∞) includes 1; ∪ joins that piece with the negative piece.
Unit-circle coordinate bounds. A circle of radius 1 never has a horizontal coordinate 2, so cos x ≠ 2.
Vertical-axis circle reflection. Reflect 60° to 120°: cos( $2 π over 3$ ) = − $1 over 2$ while sine stays $square root of 3 over 2$ .

Why it works. Sine and cosine take every value from −1 to 1. A positive value at most 1 has a reciprocal at least 1; a negative value of magnitude at most 1 has a reciprocal at most −1. Every height in those outside bands occurs: its reciprocal is an available circle coordinate. For tangent output m, point toward (1, m) and shrink both coordinates equally to the unit circle. The nonzero horizontal coordinate ensures an allowed angle, and the ratio stays m. For cotangent use (m, 1); its nonzero vertical coordinate ensures an allowed angle and ratio m.

RuleRule: tan x and cot x have range (−∞, ∞). sec x and csc x have range (−∞, −1] ∪ [1, ∞), meaning y ≤ −1 or y ≥ 1.

The same idea, five ways

Say it

Say range means possible output heights.

Write it

The range lists outputs, while the domain lists inputs.

In math

range(tan) = range(cot) = (−∞, ∞)
range(sec) = range(csc) = (−∞, −1] ∪ [1, ∞)
graph words: look at the shadow on the y-axis

A side lamp makes the graph's height shadow on a wall.

See it

Inputs run horizontally in radians; outputs run vertically.

The same idea, other ways

As a shadow

Range is the graph's shadow on the vertical axis. The secant shadow skips the middle but includes its two edges.

Inputs run horizontally in radians; outputs run vertically.

With pieces

If one piece is $1 over 4$ of a whole, four pieces make one. Taking a reciprocal of a smaller positive fraction makes a larger positive number.

One fourth becomes four under the reciprocal operation.

From a ramp

A ramp can have any finite rise per unit run. Tangent therefore has no forbidden real output heights, even though it has forbidden angle inputs.

The direction through (1, 3) has ratio 3; any other real rise can be chosen too.

.1Tangent and cotangent: every real output

Their input lists have holes, but their output lists do not. A ramp's rise can be small, large, negative or zero. Each valid ratio describes a direction on the circle. By choosing that direction you can produce the desired tangent or cotangent height.

Rule: range(tan) = range(cot) = (−∞, ∞).
Amplitude means a finite half-height of a bounded wave, half the difference between its highest and lowest outputs.
Tangent and cotangent have no amplitude because they have neither a highest nor a lowest output.

Inputs run horizontally in radians; outputs run vertically.

The same idea, five ways

Say it

Say every real output height is possible. Say no amplitude as no finite half-height.

Write it

There is no upper or lower limit on the quotient outputs. Without a highest and lowest output, a finite amplitude cannot be formed.

In math

y ∈ (−∞, ∞)
{y | y is real}
graph words: the shadow covers the entire vertical axis
amplitude = $highest output − lowest output over 2$ , when both finite heights exist

A ramp can have any finite slope.

See it

Inputs run horizontally in radians; outputs run vertically.

Worked exampleZero is an allowed quotient height

In plain words, find one permitted input for each zero. Show that tangent and cotangent can both have output 0.

Both columns show a permitted zero output.

tan 0 = $0 over 1$ = 0.Sine is zero while cosine is nonzero.
cot( $π over 2$ ) = $0 over 1$ = 0.Cosine is zero while sine is nonzero.

Answer

Tangent gives 0 at x = 0.
Cotangent gives 0 at x = $π over 2$ .

Check Both numerator-zero inputs have a bottom of 1, so neither division is forbidden.

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

✗ Not this: Counterexample: tangent has amplitude 1 because its plotted points include ±1.

Those two points are sample heights, not highest and lowest heights.

✓ Instead: Tangent reaches past every finite height, so it has no amplitude.

Tips and tricks

Tip: Do not confuse the domain's missing inputs with the range's output heights.

.2Secant and cosecant: two closed output bands

A reciprocal of a nonzero sine or cosine cannot be smaller than one in size. Picture pieces of at most one whole: you need at least one piece to make a whole. Negative pieces give the same size restriction below zero. Their range keeps two outside pieces and removes the middle.

Rule: y ≤ −1 or y ≥ 1.
Both endpoints ±1 occur.
The excluded band is (−1, 1).
Secant and cosecant also have no finite amplitude because their outside heights are unbounded.

The closed edge heights are included, with no end in either direction.

The same idea, five ways

Say it

Say at most negative one or at least one.

Write it

The reciprocal graphs occupy two separated height bands.

In math

y ∈ (−∞, −1] ∪ [1, ∞)
{y | y ≤ −1 or y ≥ 1}
graph words: no points in the strip −1 < y < 1

Two permitted roads on opposite sides of a closed middle stretch.

See it

Inputs run horizontally in radians; outputs run vertically.

Worked exampleBoth range endpoints occur

In plain words, prove the two edge heights belong to its range. Find inputs giving cosecant outputs 1 and −1.

Inputs run horizontally in radians; outputs run vertically.

csc( $π over 2$ ) = $1 over 1$ = 1.The sine at the top of the unit circle is 1.
csc( $3 π over 2$ ) = $1 over −1$ = −1.The sine at the bottom of the unit circle is −1.

Answer

1 occurs at $π over 2$ .
−1 occurs at $3 π over 2$ .

Check Multiplying each cosecant by its sine gives 1, confirming both reciprocal outputs.

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

✗ Not this: Counterexample: sec x = − is possible because it is negative.

The condition requires at most −1, not every negative number.

✓ Instead: − is in the excluded middle band.

Tips and tricks

Tip: Keep the brackets on −1 and 1 when writing the reciprocal range.

Strategy: step by step

1. Identify whether you have a quotient function or a reciprocal-of-sine-or-cosine function.
2. For tangent or cotangent, accept any real output because every height occurs.
3. For secant or cosecant, keep outputs at or below −1 and at or above 1.
4. Include −1 and 1; exclude only the open middle band −1 < y < 1.

Strategy

Strategy: check a proposed output

Is the function tangent or cotangent?

YesEvery real candidate output is allowed.

NoFor secant or cosecant check y ≤ −1 or y ≥ 1.

↓

Does a reciprocal candidate lie strictly between −1 and 1?

YesReject it; that height is in the missing middle.

NoKeep it, including ±1.

1. Read which function the output belongs to.
2. Apply the all-real quotient range or the two-band reciprocal range.
3. For a reciprocal candidate, optionally flip it back and check the required sine or cosine.
4. Include allowed endpoints and explain rejections.

Worked exampleWhich proposed secant heights are possible?

In plain words, check four proposed heights against the range. Decide whether sec x can equal 1, −2, 0 or $1 over 2$ .

The two allowed pieces contain 1 and −2, while the middle candidates lie in the gap.

Keep 1 because sec 0 = $1 over cos 0$ = $1 over 1$ = 1.The endpoint 1 is included, and input 0 gives it.
Keep −2 because cos( $2 π over 3$ ) = − $1 over 2$ , so sec( $2 π over 3$ ) = 1 ÷ (− $1 over 2$ ) = −2.A reciprocal retains the negative sign. Reflection of the 60° circle point into the upper-left region changes cosine to − $1 over 2$ .
Reject 0 because 1 divided by a finite nonzero number cannot be zero.If the reciprocal equaled 0, multiplying by its denominator would give 1 = 0.
Reject $1 over 2$ because it would require cos x = 2, outside cosine's possible coordinates.Taking the reciprocal back checks whether the proposed output could have a valid denominator.

Answer

1: possible.
−2: possible.
0: impossible.
$1 over 2$ : impossible.

Check The permitted heights satisfy y ≤ −1 or y ≥ 1; the two rejected heights lie in the gap.

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

✗ Not this: Counterexample: the range of secant is (−∞, −1) ∪ (1, ∞).

Parentheses throw away −1 and 1, even though both are actual reciprocal outputs.

✓ Instead: Use (−∞, −1] ∪ [1, ∞).

Tips and tricks

Tip: For a proposed reciprocal output, flip it back. If the required sine or cosine is beyond ±1, reject it.

Trap. Excluding ±1 along with the middle band. The gap is open: −1 < y < 1. At the edges, reciprocal values are still ±1.

Keep in mind

The edges count: csc x reaches 1 at the tops of the sine wave and −1 at its bottoms, so write [1, ∞) and (−∞, −1] with square brackets.
A graph window cuts off the picture, not the function: on a screen showing −4 ≤ y ≤ 4, tan x still reaches 100 above the top edge.
The graph of csc x never touches the x-axis: 1 divided by a number is never 0, even 1 ÷ 10000 = 0.0001.

Memory hookSec and csc skip the middle band −1 < y < 1 but touch its edges, ±1; tan and cot reach every height.

Flash cards: say the answer out loud, then flip

What is the range of a function?

All its outputs: every height its graph reaches.

Range of cot x?

(−∞, ∞): every real number.

Range of csc x?

(−∞, −1] ∪ [1, ∞).

Can csc x = ?

No: sin x would be ≈ 1.33, bigger than 1.

Can sec x = −1.6?

Yes: cos x would be 1 ÷ (−1.6) = −0.625, inside [−1, 1].

Why is the range of sec x not written (−∞, −1) ∪ (1, ∞)?

Because sec x does reach −1 and 1, at the valleys and peaks of cos x, so both ends need square brackets.