Even and odd: what a minus sign inside does

Explain it like I am five

Picture a tree beside a still lake. Its reflection has the same left-right position but is upside down. On the unit circle a negative angle −t (read 'negative t') walks the same distance clockwise instead of counterclockwise, and it lands on the reflection of the point for t across the x-axis.

Example: say t ends at the point (0.28, 0.96). Then −t ends at (0.28, −0.96): same x, opposite y. Cosine reads x, so cos(−t) = 0.28 = cos t. Sine reads y, so sin(−t) = −0.96 = −sin t.

A function f is even when a minus sign inside disappears: f(−t) = f(t), read 'f of negative t equals f of t', like cosine. It is odd when the minus sign comes out front: f(−t) = −f(t), like sine. Secant is 1 divided by cosine, so it is even too. Tangent, cotangent and cosecant each hold exactly one sine, so each flips sign: they are odd.

In plain words

Imagine walking around a circular track. You can walk the same distance forward or backward. The two stopping places sit directly above and below each other. Their left or right positions match, but their heights have opposite signs. A negative input −t means reversing the direction of the walk. A function is even when that reversal keeps its output the same. It is odd when the reversal changes the output to its negative. Cosine measures left or right position, so cosine is even. Sine measures height, so sine is odd. The four functions built by dividing or taking reciprocals follow from those two facts. In either definition, the comparison must hold for every allowed input.

Cosine is even: t = and t = − give the same height, , so the graph is a mirror image across the y-axis.

Reminder

Negative of a negative number. An outside negative sign reverses the output: −(− $1 over 2$ ) = $1 over 2$ .
Periods. Remove whole 2π periods for sine, cosine, secant and cosecant. Tangent and cotangent allow whole π periods.
Reciprocals. Flipping a nonzero fraction keeps its sign: 1 ÷ (− $1 over 3$ ) = −3.

Cosine is even: t = and t = − give the same height, , so the graph is a mirror image across the y-axis.

Read the output entry in the column under its input or category.

Why it works. If P(t) = (a, b), reflection across the x-axis gives P(−t) = (a, −b). Thus cos(−t) = a = cos t and sin(−t) = −b = −sin t. Dividing the opposite height by the unchanged horizontal coordinate reverses tangent's sign. Dividing the unchanged coordinate by the opposite height reverses cotangent's sign. Taking a reciprocal preserves cosine's equality and sine's sign reversal. Each statement applies where the function is defined; reflection preserves those allowed inputs.

RuleAn even or odd function has a domain that contains −t whenever it contains t. For every t in that domain, even means f(−t) = f(t), and odd means f(−t) = −f(t). Cosine and secant are even. Sine, cosecant, tangent and cotangent are odd. Even graphs are symmetric about the y-axis, meaning the two sides match when folded across that axis. Odd graphs are symmetric about the origin, meaning a half turn around the origin leaves the graph unchanged.

The same idea, five ways

Say it

An even function gives the same output for opposite inputs. An odd function gives opposite outputs for opposite inputs.

Write it

Reflecting an even function's graph across the y-axis keeps it unchanged; turning an odd function's graph half a turn about the origin keeps it unchanged.

In math

Even: f(−t) = f(t) for every input t in a domain symmetric about 0.
Odd: f(−t) = −f(t) for every input t in a domain symmetric about 0.
Even: cos t, sec t.
Odd: sin t, csc t, tan t, cot t.

Walk the same distance forward and backward on a circular track. Compare the two stopping places.

See it

Cosine has equal outputs at opposite inputs.

The same idea, other ways

As two walks

Walk t around the circle from the rightmost point. Then walk the same distance backward. You reach (a, b) and (a, −b). Matching horizontal positions explain cosine; opposite heights explain sine.

Walking backward reflects the stopping point across the x-axis.

As graph symmetry

An even graph contains (t, f(t)) and (−t, f(t)). Folding along the y-axis matches those points. An odd graph contains (t, f(t)) and (−t, −f(t)). A half turn about the origin matches those points. These graph pictures follow from the definitions, rather than giving new sign rules.

Sine's paired graph points have opposite input and output signs.

With the fraction definitions

Start with sin(−t) = −sin t and cos(−t) = cos t. Then tan(−t) = $−sin t over cos t$ = −tan t, while sec(−t) = $1 over cos t$ = sec t. The same substitutions give odd cotangent and cosecant.

sin(−t) = −sin t

cos(−t) = cos t

tan(−t) = −tan t

sec(−t) = sec t

Only cosine and its reciprocal keep the output unchanged.

Function	f(−t) equals	Reason	Type
sin t	−sin t	P(−t) has the opposite height	odd
cos t	cos t	P(−t) has the same x	even
tan t	−tan t	$sin(−t) over cos(−t)$ = $−sin t over cos t$	odd
cot t	−cot t	$cos(−t) over sin(−t)$ = $cos t over −sin t$	odd
sec t	sec t	$1 over cos(−t)$ = $1 over cos t$	even
csc t	−csc t	$1 over sin(−t)$ = $1 over −sin t$	odd

.1Sine is odd

Sine reads the stopping point's height. Reversing the walk reflects that height across 0.

sin(−t) = −sin t.
Sine is an odd function.
Its graph is symmetric about the origin.

Sine's graph matches itself after a half turn about the origin.

Worked exampleSine at a backward quarter turn

The angle is −90°. Find the sine, which is the stopping point's height.

The height is −1 after a clockwise quarter turn.

sin(−90°) = −sin 90°.Sine is odd.
sin 90° = 1, so sin(−90°) = −1.The top of the unit circle has height 1.

Answer

−1

Check A clockwise quarter turn reaches (0, −1), whose height is −1.

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

✗ Not this: sin(−t) = sin t.

The reflected point has the opposite height, unless that height is 0.

✓ Instead: sin(−t) = −sin t.

Tips and tricks

For sine, reverse the height's sign.

.2Cosine is even

Cosine reads left or right position. Reversing the walk preserves that horizontal position.

cos(−t) = cos t.
Cosine is an even function.
Its graph is symmetric about the y-axis.

Cosine reflects onto itself across the y-axis.

Worked exampleCosine at opposite sixth-turn inputs

The input is − $π over 3$ . Find its cosine.

The negative angle has the same x-coordinate as the positive angle.

cos(− $π over 3$ ) = cos $π over 3$ .Cosine is even.
cos $π over 3$ = $1 over 2$ .The 60° special triangle gives x = $1 over 2$ on the unit circle.

Answer

Check The two points have coordinates (, ) and (, −). Their x-coordinates match.

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

✗ Not this: cos(−) = −.

The minus sign belongs to the input. Cosine is even, so its output does not reverse.

✓ Instead: cos(−) = .

Tips and tricks

For cosine, preserve the horizontal position.

.3Tangent is odd

Tangent divides height by horizontal position. Reversing only the height reverses the quotient.

tan(−t) = $−sin t over cos t$ = −tan t.
Both sides require cos t ≠ 0.

Tangent's paired values have opposite signs wherever both are defined.

Worked exampleTangent at a negative special angle

The input is − $π over 6$ . Find its exact tangent.

The quotient has a negative top and a positive bottom.

tan(− $π over 6$ ) = −tan $π over 6$ .Tangent is odd.
tan $π over 6$ = $1 over square root of 3$ = $square root of 3 over 3$ , so the answer is − $square root of 3 over 3$ .Opposite ÷ adjacent in the 30° triangle gives $1 over square root of 3$ , and multiplying top and bottom by $square root of 3$ rationalizes it.

Answer

−

Check The original angle is in Quadrant IV. Its sine is negative and its cosine is positive, so the tangent must be negative.

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

✗ Not this: tan(−t) = tan t because tangent repeats.

Repeating by π and reflecting the input are different operations. At t = , reflection changes 1 to −1.

✓ Instead: tan(−t) = −tan t; tan(t + π) = tan t.

Tips and tricks

Keep odd symmetry separate from period.

.4Cotangent is odd

Cotangent divides horizontal position by height. Reversing the height puts one negative sign on the bottom, which reverses the quotient.

cot(−t) = $cos t over −sin t$ = −cot t.
Both sides require sin t ≠ 0.

Cotangent has opposite outputs at opposite allowed inputs.

Worked exampleCotangent at a negative eighth-turn input

The input is − $π over 4$ . Find its cotangent.

Equal leg sizes with different signs give cotangent −1.

cot(− $π over 4$ ) = −cot $π over 4$ .Cotangent is odd.
cot $π over 4$ = cos $π over 4$ ÷ sin $π over 4$ = 1, so cot(− $π over 4$ ) = −1.The 45° special triangle has equal legs, so their quotient is 1.

Answer

−1

Check At −45°, the coordinates have equal magnitudes with x positive and y negative. The quotient x ÷ y is −1.

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

✗ Not this: A negative sign in the bottom disappears.

A positive divided by a negative is negative; there is only one sign reversal.

✓ Instead: = − when b ≠ 0.

Tips and tricks

A negative sign on either side of a fraction gives the fraction one negative sign.

.5Secant is even

Secant takes the reciprocal of horizontal position. Since that position stays unchanged, its reciprocal stays unchanged too.

sec(−t) = $1 over cos(−t)$ = $1 over cos t$ = sec t.
Both sides require cos t ≠ 0.

Secant is a mirror image across the y-axis.

Worked exampleSecant at a negative angle beyond one turn

The input is − $7 π over 3$ . Find its secant.

The positive input reduces from 420° to 60°.

sec(− $7 π over 3$ ) = sec $7 π over 3$ .Secant is even.
$7 π over 3$ − 2π = $π over 3$ , so sec $7 π over 3$ = sec $π over 3$ .Secant has period 2π.
sec $π over 3$ = 1 ÷ cos $π over 3$ = 1 ÷ $1 over 2$ = 2.Secant is the reciprocal of cosine.

Answer

Check 2 is in secant's range and is positive, as the original Quadrant IV angle requires.

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

✗ Not this: sec(−t) = −sec t.

Secant inherits cosine's even symmetry, rather than sine's odd symmetry.

✓ Instead: sec(−t) = sec t.

Tips and tricks

Remember cosine and secant as the even pair.

.6Cosecant is odd

Cosecant takes the reciprocal of height. Reversing the height reverses the reciprocal's sign.

csc(−t) = $1 over −sin t$ = −csc t.
Both sides require sin t ≠ 0.

Cosecant has origin symmetry wherever it is defined.

Worked exampleCosecant at a negative special angle

The input is − $π over 6$ . Find its cosecant.

Taking the reciprocal of the negative height gives negative cosecant.

csc(− $π over 6$ ) = −csc $π over 6$ .Cosecant is odd.
csc $π over 6$ = 1 ÷ $1 over 2$ = 2, so the answer is −2.Sine at $π over 6$ is $1 over 2$ , and cosecant is its reciprocal.

Answer

−2

Check At the original input, sin(−) = −. Its reciprocal is −2 directly.

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

✗ Not this: csc(−t) = csc t.

The reciprocal keeps the sign of sine, so it reverses when sine reverses.

✓ Instead: csc(−t) = −csc t.

Tips and tricks

Cosecant follows sine, including its sign.

Strategy: step by step

1. Read the negative sign inside the function as an instruction to reverse the input.
2. Identify the function. Cosine and secant are even. Sine, cosecant, tangent and cotangent are odd.
3. For an even function, replace f(−t) by f(t). For an odd function, replace f(−t) by −f(t). Write the outside sign immediately.
4. Evaluate the positive input using a known value. If it still contains extra periods, use the period already taught to reduce it.
5. Keep the outside sign through every later step. Check that the resulting sign agrees with the terminal point.
Read a reference-table entry in the column under its input or category in the matching picture.

Strategy

Evaluate a function at a negative input

Is the function cosine or secant?

YesUse f(−t) = f(t). Keep the positive-input output, including any sign it already has.

NoFor sine, cosecant, tangent or cotangent, use f(−t) = −f(t). Keep the outside negative sign.

↓

Is the resulting positive angle one of the known exact angles?

YesRead its exact value, then apply the outside sign.

NoReduce by whole periods if possible. If it is still unfamiliar, give a sign or an equivalent expression rather than inventing an exact special value.

1. Read the negative sign inside the function as an instruction to reverse the input.
2. Identify the function. Cosine and secant are even. Sine, cosecant, tangent and cotangent are odd.
3. For an even function, replace f(−t) by f(t). For an odd function, replace f(−t) by −f(t). Write the outside sign immediately.
4. Evaluate the positive input using a known value. If it still contains extra periods, use the period already taught to reduce it.
5. Keep the outside sign through every later step. Check that the resulting sign agrees with the terminal point.

Worked exampleUsing even and odd identities to evaluate functions of − and −

The terminal point of t = $5 π over 6$ on the unit circle is P(− $square root of 3 over 2$ , $1 over 2$ ). Use the even and odd properties of the trigonometric functions to find the exact value of each expression.
(a) sin(− $5 π over 6$ )
(b) sec(− $5 π over 6$ )
(c) cot(− $5 π over 6$ )
(d) csc(− $17 π over 6$ )

Unit circle with the terminal point P(−, ) of t = (150°). Reversing the input reflects P across the x-axis to (−, −).

Read each minus sign inside the function as reversing the input. A reversed input moves the terminal point from P(− $square root of 3 over 2$ , $1 over 2$ ) to its reflection across the x-axis, (− $square root of 3 over 2$ , − $1 over 2$ ).Going around the unit circle clockwise instead of counterclockwise by the same amount reflects the point across the x-axis. The x-coordinate stays the same and the y-coordinate changes sign.
Identify each function. Sine, cotangent and cosecant are odd. Secant is even.Cosine and secant are the only even functions among the six. Sine, cosecant, tangent and cotangent are odd.
(a) Write sin(− $5 π over 6$ ) = −sin( $5 π over 6$ ). The y-coordinate of P gives sin( $5 π over 6$ ) = $1 over 2$ . So sin(− $5 π over 6$ ) = − $1 over 2$ .Sine is odd, so f(−t) = −f(t). The outside sign is written right away and kept.
(b) Write sec(− $5 π over 6$ ) = sec( $5 π over 6$ ) = $\frac{1}{\cos (\frac{5 π}{6})}$ . This is $\frac{1}{- \frac{\sqrt{3}}{2}}$ = − $2 over square root of 3$ = − $2 square root of 3 over 3$ .Secant is even, so f(−t) = f(t) and no sign is added. Its value is the reciprocal of the x-coordinate of P. The denominator is then rationalized.
(c) Write cot(− $5 π over 6$ ) = −cot( $5 π over 6$ ). From P, cot( $5 π over 6$ ) = $\frac{\cos t}{\sin t}$ = $\frac{- \frac{\sqrt{3}}{2}}{\frac{1}{2}}$ = − $square root of 3$ . So cot(− $5 π over 6$ ) = −(− $square root of 3$ ) = $square root of 3$ .Cotangent is odd, so the outside minus sign is written first. It is then kept while the inside value, which is itself negative, is evaluated.
(d) Write csc(− $17 π over 6$ ) = −csc( $17 π over 6$ ). Since $17 π over 6$ = 2π + $5 π over 6$ and cosecant has period 2π, csc( $17 π over 6$ ) = csc( $5 π over 6$ ) = $\frac{1}{\frac{1}{2}}$ = 2. So csc(− $17 π over 6$ ) = −2.Cosecant is odd, so the outside sign comes first. Next the extra full period 2π is removed, which leaves the known input $5 π over 6$ .
Check the signs against the terminal points. − $5 π over 6$ ends in Quadrant III at (− $square root of 3 over 2$ , − $1 over 2$ ). Also, − $17 π over 6$ + 4π = $7 π over 6$ , which is also in Quadrant III.In Quadrant III, sine, cosecant, cosine and secant are negative, and tangent and cotangent are positive. This matches − $1 over 2$ , − $2 square root of 3 over 3$ , $square root of 3$ and −2.

Answer

(a) sin(−) = − (b) sec(−) = − (c) cot(−) = (d) csc(−) = −2

Check Read the values straight from the reflected point (−, −) for −. sin = −. sec = = −. cot = = . The angle − is coterminal with , whose point is (−, −), so csc = = −2. All four values agree with the answers.

Work to write

sin is odd: sin(− $5 π over 6$ ) = −sin( $5 π over 6$ ) = − $1 over 2$
sec is even: sec(− $5 π over 6$ ) = sec( $5 π over 6$ ) = $\frac{1}{- \frac{\sqrt{3}}{2}}$ = − $2 square root of 3 over 3$
cot is odd: cot(− $5 π over 6$ ) = −cot( $5 π over 6$ ) = −(− $square root of 3$ ) = $square root of 3$
csc is odd: csc(− $17 π over 6$ ) = −csc( $17 π over 6$ ) = −csc( $5 π over 6$ ) = −2, using period 2π
Sign check: − $5 π over 6$ and − $17 π over 6$ both end in Quadrant III, where sin, csc and sec are negative and cot is positive

(a) sin(− $5 π over 6$ ) = − $1 over 2$ (b) sec(− $5 π over 6$ ) = − $2 square root of 3 over 3$ (c) cot(− $5 π over 6$ ) = $square root of 3$ (d) csc(− $17 π over 6$ ) = −2

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

Rung 1Evaluating cosine at a negative angle using evenness

Find the exact value of cos(− $3 π over 4$ ) using the fact that cosine is an even function.

Unit circle with the terminal point P of − (the same position as 225°) at (−, −). Its x-coordinate, −, is the value of cos(−).

Read cos(− $3 π over 4$ ) as cosine evaluated at the input $3 π over 4$ reversed. The angle is measured $3 π over 4$ clockwise from the positive x-axis.A negative sign inside the function reverses the input. It turns the direction of rotation from counterclockwise to clockwise.
Identify the function as cosine, which is even.Cosine and secant are the even trigonometric functions. Sine, cosecant, tangent and cotangent are odd.
Replace cos(− $3 π over 4$ ) by cos( $3 π over 4$ ). No minus sign is placed outside.For an even function f(−t) = f(t), so reversing the input leaves the output unchanged.
Evaluate cos( $3 π over 4$ ). The angle $3 π over 4$ lies in Quadrant II with reference angle $π over 4$ , and its terminal point is (− $square root of 2 over 2$ , $square root of 2 over 2$ ). So cos( $3 π over 4$ ) = − $square root of 2 over 2$ . No period reduction is needed because $3 π over 4$ is already between 0 and 2π.Cosine is the x-coordinate of the terminal point. The known value at $π over 4$ is $square root of 2 over 2$ , and x is negative in Quadrant II.
Check the sign against the original angle. − $3 π over 4$ ends in Quadrant III at (− $square root of 2 over 2$ , − $square root of 2 over 2$ ). Its x-coordinate is negative, which agrees with − $square root of 2 over 2$ .The sign of the result must match the terminal point of the actual input. Evenness changed no sign, so the negative value comes from the quadrant alone.

Answer

cos(−) = −

Check Reflecting the point P for − at (−, −) across the x-axis gives (−, ), which is the point for . The x-coordinate is unchanged, as an even function requires. Both points have x = −.

Work to write

cosine is even, so cos(−t) = cos(t)
cos(− $3 π over 4$ ) = cos( $3 π over 4$ )
$3 π over 4$ is in Quadrant II with reference angle $π over 4$ , so cos( $3 π over 4$ ) = − $square root of 2 over 2$
cos(− $3 π over 4$ ) = − $square root of 2 over 2$
check: − $3 π over 4$ ends in Quadrant III, where x < 0

cos(− $3 π over 4$ ) = − $square root of 2 over 2$

Rung 2Rung 2: an odd function

Find sin(−30°). In words, reverse the known sine output.

The original angle has negative height.

sin(−30°) = −sin 30°.Sine is odd.
−sin 30° = − $1 over 2$ .The known sine at 30° is $1 over 2$ .

Answer

−

Check The angle is below the x-axis, so its sine is negative.

Rung 3Cotangent of a reversed angle with extra periods

The terminal point of t = $π over 3$ on the unit circle is P( $1 over 2$ , $square root of 3 over 2$ ). Use the fact that cotangent is an odd function, together with its period π, to find the exact value of cot(− $19 π over 3$ ). Rationalize the denominator.

Unit circle with the terminal point P(, ) of t = , which is used to evaluate cot().

Read cot(− $19 π over 3$ ) as cotangent evaluated at the input $19 π over 3$ taken in the reverse (clockwise) direction.A negative sign inside the function reverses the input. The negative sign does not yet tell us the sign of the output.
Identify the function as cotangent, which is odd.Sine, cosecant, tangent and cotangent are odd, so f(−t) = −f(t).
Write cot(− $19 π over 3$ ) = −cot( $19 π over 3$ ).For an odd function the minus sign moves outside. We write it immediately so that it is not lost.
Split the input: $19 π over 3$ = $18 π over 3$ + $π over 3$ = 6π + $π over 3$ .The remaining input still contains extra periods, and these must be removed before we use a known value.
Reduce: cot( $19 π over 3$ ) = cot( $π over 3$ + 6π) = cot( $π over 3$ ).Cotangent has period π, and 6π is six whole periods. Adding whole periods does not change the value.
Evaluate cot( $π over 3$ ) = $\frac{x}{y}$ = $1/2 over square root of 3 /2$ = $1 over square root of 3$ = $square root of 3 over 3$ .Cotangent is the quotient x/y of the terminal point coordinates, and P( $1 over 2$ , $square root of 3 over 2$ ) is the given terminal point for $π over 3$ . Multiplying the top and bottom by $square root of 3$ rationalizes the denominator.
Restore the outside sign: cot(− $19 π over 3$ ) = − $square root of 3 over 3$ .The minus sign from the odd property has to be kept through every later step.

Answer

cot(−) = −

Check − = −6π − , so the angle is coterminal with −. Its terminal point is the reflection of P across the x-axis, which is (, −) in Quadrant IV. In Quadrant IV, x > 0 and y < 0, so cotangent is negative. The direct quotient is x/y = = − = −. This matches the answer.

Work to write

cot is odd, so cot(− $19 π over 3$ ) = −cot( $19 π over 3$ )
$19 π over 3$ = 6π + $π over 3$ and the period of cot is π, so cot( $19 π over 3$ ) = cot( $π over 3$ )
cot( $π over 3$ ) = $1/2 over square root of 3 /2$ = $square root of 3 over 3$
cot(− $19 π over 3$ ) = − $square root of 3 over 3$

cot(− $19 π over 3$ ) = − $square root of 3 over 3$

Rung 4Rung 4: even symmetry and a period

Find sec(− $7 π over 3$ ). In words, reverse the input and remove a whole period.

The positive input has one extra full turn.

sec(− $7 π over 3$ ) = sec $7 π over 3$ .Secant is even.
$7 π over 3$ − $6 π over 3$ = $π over 3$ .Secant's period is 2π = $6 π over 3$ .
sec $π over 3$ = 1 ÷ $1 over 2$ = 2.Cosine at $π over 3$ is $1 over 2$ .

Answer

Check The original angle reduces by full turns to , in Quadrant IV, where secant is positive.

Rung 5Rung 5: odd symmetry over several periods

Find cot(− $9 π over 4$ ). In words, preserve the outside sign while reducing the positive input.

The equal coordinate sizes have opposite signs.

cot(− $9 π over 4$ ) = −cot $9 π over 4$ .Cotangent is odd.
$9 π over 4$ − 2π = $9 π over 4$ − $8 π over 4$ = $π over 4$ .Two cotangent periods, each π, preserve its value.
−cot $π over 4$ = −1.Cosine and sine at $π over 4$ are equal and nonzero, so cotangent is 1.

Answer

−1

Check The original input is coterminal with , in Quadrant IV, where cotangent is negative.

Rung 6Rung 6: the same negative input with an even function

Find cos(− $9 π over 4$ ). In words, compare with the previous rung and notice which sign rule changes.

Cosine is positive while cotangent is negative at the same original input.

cos(− $9 π over 4$ ) = cos $9 π over 4$ .Cosine is even, so no outside minus sign is introduced.
$9 π over 4$ − 2π = $π over 4$ .Cosine has period 2π.
cos $π over 4$ = $square root of 2 over 2$ .The 45° triangle supplies the exact value.

Answer

Check The original terminal point is in Quadrant IV. Its x-coordinate is positive, matching the answer.

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

✗ Not this: cos(−t) = −cos t.

Changing the direction of the walk preserves the horizontal coordinate. At t = , cosine is at both inputs.

✓ Instead: cos(−t) = cos t. Move the minus sign outside only for an odd function.

✗ Not this: An even function always has positive outputs.

Even describes how opposite inputs compare. It does not describe the output's sign. At t = 3π, cosine is −1, and cos(−3π) = −1 too.

✓ Instead: An even function has equal outputs at opposite inputs, including equal negative outputs.

Tips and tricks

Memory device: cosine and its reciprocal, secant, are the only even trig functions. The other four are odd.
Write the sign change as an equality before computing. A negative sign inside and a negative sign outside mean different things.

Trap. Moving a minus sign outside cosine or secant changes the answer incorrectly. Even functions preserve the positive-input output's sign. Odd functions reverse that output's sign; if the positive-input output is already negative, the result is positive.

Keep in mind

A minus inside changes the angle and a minus outside flips the answer: with sin t = 0.96, sin(−t) = −0.96, but −sin(−t) = 0.96.
A negative angle mirrors across the x-axis, (x, y) to (x, −y); the y-axis mirror, (x, y) to (−x, y), belongs to a different angle, π − t (180° − t in degrees).
Even does not mean positive: cos(−3) and cos 3 are both about −0.990.
In an identity such as cos(−t) = cos t, LHS means the left-hand side, cos(−t), and RHS the right-hand side, cos t; a proof rewrites one side until it matches the other.

Memory hookCosine eats the minus sign; sine spits it out. Secant copies cosine; tangent, cotangent and cosecant copy sine.

Flash cards: say the answer out loud, then flip

What does 'odd function' mean?

f(−t) = −f(t) for every allowed t: a minus sign inside comes out front.

Which trig functions are even, and which are odd?

Even: cos, sec.
Odd: sin, csc, tan, cot.

csc t = 4. What is csc(−t)?

−4, because cosecant is odd.

P(t) = (−0.96, 0.28). Find cos(−t) and sin(−t).

The point for −t is (−0.96, −0.28).
cos(−t) = −0.96
sin(−t) = −0.28

cos t = 0.28. Is −cos t equal to cos(−t)?

No: cos(−t) = 0.28, but −cos t = −0.28.

In sin(−t) = −sin t, which side is the LHS?

sin(−t), the left-hand side of the equals sign.