Combining properties to find signs and known exact values

Explain it like I am five

Picture a car on a circular track whose trip log reads 'backward 1110 degrees'. To find where it stopped, first note that it went backward, then ignore the full laps, then look at the spot. Angles work the same way, in four steps.

Example: csc(−1110°), where csc means cosecant, 1 divided by sine.
Step 1, the minus inside: cosecant is odd (a minus inside comes out front), so csc(−1110°) = −csc 1110°. Keep that outside minus.
Step 2, whole laps: a full lap leaves the point where it was, and 1110 ÷ 360 is about 3.08, so remove 3 × 360° = 1080°, leaving 30°.
Step 3, the sign: 30° is in Quadrant I, where all six functions are positive.
Step 4, the known value: sin 30° = $1 over 2$ , so csc 30° = 1 ÷ $1 over 2$ = 2. With the saved minus, csc(−1110°) = −2.

Check another way: −1110° + 4 × 360° = 330°, in Quadrant IV, below the x-axis, where cosecant is negative. It agrees.

In plain words

Think of reading a clock after several full days have passed. You remove whole days to find the same place on the clock face. You can do that with angles too. First handle a reversed walk, then remove extra repeats, then identify the stopping place. That place tells you whether the output is positive, negative, zero or undefined. If the remaining angle is one whose exact value you have learned, you can finish with that value. Sometimes the angle is unfamiliar. Then these properties still give its sign and an equivalent smaller angle, but they do not create a special exact value for every possible angle.

750° is two full turns plus 30°, so it ends exactly where 30° ends.

Reminder

Fractions with π. Match denominators before subtracting: 2π = $8 π over 4$ , so $9 π over 4$ − 2π = $π over 4$ .
Sign multiplication. An outside negative sign times a negative output is positive: −(− $1 over 2$ ) = $1 over 2$ .
Zero in a denominator. $0 over −1$ = 0 is defined. $1 over 0$ has no value.

750° is two full turns plus 30°, so it ends exactly where 30° ends.

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

Why it works. Each replacement preserves the expression's value. Even and odd symmetry handles the negative input, with a possible negative sign outside. A whole period preserves the function's output. The quadrant or axis point identifies its sign and whether a denominator is zero. A known special angle or an identity then gives an exact value when available. An angle such as 17° cannot be reduced to 30° by removing full turns, so the method must stop with its known sign or the expression itself.

RuleHandle the input's minus sign, remove whole periods, identify a quadrant or an axis, then use a known exact value or a suitable identity. For an unfamiliar angle, the properties can determine a sign or a smaller equivalent expression without determining a special exact value.

The same idea, five ways

Say it

Find the same output by replacing a difficult angle with one you can recognize.

Write it

Symmetry and periods simplify the input; its terminal point decides the sign and possible denominator failures.

In math

sin(−t) = −sin t; cos(−t) = cos t.
sin(t + 2nπ) = sin t; tan(t + nπ) = tan t, for integer n.
At P(x, y), sin t = y, cos t = x, tan t = $\frac{y}{x}$ , cot t = $\frac{x}{y}$ , sec t = $1 over x$ , csc t = $1 over y$ .

Remove full days from a clock reading, then inspect the place where the hand stops.

See it

Removing two full turns changes 750° into 30° without changing the stopping point.

The same idea, other ways

As a sequence of equivalent expressions

Keep an equality at every stage: sin(−750°) = −sin 750° = −sin 30° = − $1 over 2$ . The outside negative sign stays visible while the input gets smaller.

sin(−750°)

= −sin 750°

= −sin 30°

The sign is carried through every equality.

As one direct walk

You may also add whole turns directly to a negative input. For example, −750° + 3 × 360° = 330°. The original stopping place is in Quadrant IV, so sine is negative. This gives an independent sign check on the odd-symmetry route.

Both rotations stop in Quadrant IV.

As the question's stopping point

A sign question ends after you know the quadrant or axis. An exact-value question needs one more ingredient, a value you know or an identity that cancels unfamiliar values. Do not keep computing after you have answered the requested question.

Sign: locate and combine signs.

Exact value: use a known value or identity.

On an axis: inspect denominators.

The question determines which final step is needed.

t	sin t	cos t	tan t	cot t	sec t	csc t
0 (0°)	0	1	0	undefined	1	undefined
$π over 6$ (30°)	$1 over 2$	$square root of 3 over 2$	$square root of 3 over 3$	$square root of 3$	$2 square root of 3 over 3$	2
$π over 4$ (45°)	$square root of 2 over 2$	$square root of 2 over 2$	1	1	$square root of 2$	$square root of 2$
$π over 3$ (60°)	$square root of 3 over 2$	$1 over 2$	$square root of 3$	$square root of 3 over 3$	2	$2 square root of 3 over 3$
$π over 2$ (90°)	1	0	undefined	0	undefined	1
π (180°)	0	−1	0	undefined	−1	undefined
$3 π over 2$ (270°)	−1	0	undefined	0	undefined	−1

.1First decide what the question asks

A question may ask for a sign rather than a full value, the way a map may ask whether a town lies east or west without asking its distance.

A sign answer is positive, negative or zero.
An undefined expression has no sign or output.
An exact value needs more information than a quadrant alone.

Sign: positive, negative or zero.

Undefined: no output.

Exact value: preserve the exact amount.

State the type of answer the question requests.

Worked exampleAn unfamiliar angle can still have a known sign

Decide the sign of sin 17°. The question asks for positive, negative or zero.

The positive height settles the sign question.

0° < 17° < 90°, so the terminal point is in Quadrant I.This is between the rightmost and topmost axis points.
sin 17° > 0.Sine is the positive height in Quadrant I.

Answer

positive

Check The point is above the x-axis. Its precise height is unnecessary for a sign question.

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

✗ Not this: The sine in Quadrant I must be .

A quadrant contains many heights; belongs to specific angles.

✓ Instead: Quadrant I tells you sine is positive, not which positive value it has.

Tips and tricks

Underline the requested answer type before working.

.2Remove a negative input with symmetry

Reverse a backward walk into a forward walk, while recording whether the function reverses its output.

Cosine and secant preserve the output.
Sine, cosecant, tangent and cotangent reverse the output.

Even: f(−t) = f(t).

Odd: f(−t) = −f(t).

Choose the function's symmetry before reducing.

Worked exampleKeep the outside sign visible

Rewrite sin(−750°) using a positive input before evaluating it.

The backward walk ends below the x-axis.

sin(−750°) = −sin 750°.Sine is odd, so reversing the input reverses the output.

Answer

−sin 750°

Check The original input reduces to 330°, where sine is negative, so the eventual outside sign must be kept.

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

✗ Not this: Drop every input minus sign.

Only even functions allow that.

✓ Instead: For odd functions, move the minus sign outside.

Tips and tricks

Write the outside negative sign before changing the angle.

.3Remove repeats with the correct period

Remove complete repeats the way you remove full days from a clock reading.

Sine, cosine, secant and cosecant repeat every 2π.
Tangent and cotangent repeat every π.
A π shortcut may change the quadrant while preserving a tangent or cotangent value.

A full-turn reduction preserves every function's value.

Worked exampleA tangent period is shorter than a full turn

Rewrite tan $13 π over 4$ with a known input.

Both coordinates reverse together, preserving tangent.

3π = $12 π over 4$ , so $13 π over 4$ − 3π = $π over 4$ .Three tangent periods fit inside the positive input.
tan $13 π over 4$ = tan $π over 4$ = 1.Tangent repeats every π.

Answer

Check The full-turn representative is , in Quadrant III, where tangent is also positive.

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

✗ Not this: Subtract π from every trig input.

For sine, cosine, secant and cosecant a half turn changes the sign.

✓ Instead: Use π only for tangent and cotangent; use 2π for the other four.

Tips and tricks

Write the function's period beside the problem.

.4Use the stopping point and a known value

After removing repeats, inspect where the walk ends. A familiar point gives an exact coordinate. An axis point also reveals denominator zeros.

A quadrant gives signs.
A known special angle gives exact coordinates.
An axis point can give zero outputs or undefined fractions.

Inspect the denominator at an axis point.

Worked exampleAn axis prevents evaluation of a reciprocal

Decide whether sec(−810°) has a value. The question asks whether cosine's reciprocal can be taken.

A horizontal coordinate of 0 makes secant undefined.

Secant is even, so consider the positive input 810°. Remove two full turns: 810° − 720° = 90°.Secant preserves opposite inputs and repeats every 360°.
At 90°, P = (0, 1), so cos 90° = 0.A counterclockwise quarter turn reaches the top point.
Secant would require dividing 1 by 0, so the original expression is undefined.A zero denominator has no quotient.

Answer

undefined

Check Direct reduction gives −810° + 3 × 360° = 270°, whose cosine is also 0.

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

✗ Not this: sec 90° = 0 because cos 90° = 0.

Taking a reciprocal puts that 0 on the bottom.

✓ Instead: sec 90° is undefined.

Tips and tricks

Use the coordinate first, then build the requested function.

Strategy: step by step

1. Read whether the request is a location or a function value or sign. For a location, keep the original signed angle, reduce only by full turns of 2π or 360°, then use its terminal point and stop. Changing the sign of the angle or using a tangent half-period may change its location.
2. If the angle is negative, use even or odd symmetry and record any outside negative sign.
3. Remove whole periods: 2π or 360° for sin, cos, sec and csc; π or 180° for tan and cot.
4. Locate the reduced angle. If it is on an axis, use its coordinates and check denominators. Otherwise use the quadrant to determine the reduced expression's sign.
5. For a sign question, combine the reduced expression's sign with the outside sign and stop. For an exact-value question, use a known value or an identity. If neither applies, retain the exact expression instead of inventing a value.
Read a reference-table entry in the column under its input or category in the matching picture.

Strategy

Reduce and evaluate a trig expression

Does the question ask for the original angle's location?

YesReduce the original signed angle by full turns of 2π or 360°. Name its original quadrant or axis and stop.

NoContinue with even or odd symmetry and the requested function's period to find its value or sign.

↓

Is the original input negative?

YesApply even or odd symmetry and record the outside sign.

NoKeep the expression's present outside sign.

↓

Is the reduced angle on an axis?

YesUse the axis coordinates. A denominator of 0 means undefined, even if an outside sign is present.

NoUse the quadrant sign rules.

↓

Does the question ask only for the sign?

YesCombine the outside sign and the reduced expression's sign, then stop.

NoLook for a known special value or an identity.

↓

Is there a known exact value or an identity that finishes the calculation?

YesShow that substitution or simplification and give the exact result.

NoGive the equivalent exact expression and any known sign. These properties alone do not supply every unfamiliar value.

1. Read whether the request is a location or a function value or sign. For a location, keep the original signed angle, reduce only by full turns of 2π or 360°, then use its terminal point and stop. Changing the sign of the angle or using a tangent half-period may change its location.
2. If the angle is negative, use even or odd symmetry and record any outside negative sign.
3. Remove whole periods: 2π or 360° for sin, cos, sec and csc; π or 180° for tan and cot.
4. Locate the reduced angle. If it is on an axis, use its coordinates and check denominators. Otherwise use the quadrant to determine the reduced expression's sign.
5. For a sign question, combine the reduced expression's sign with the outside sign and stop. For an exact-value question, use a known value or an identity. If neither applies, retain the exact expression instead of inventing a value.

Worked exampleOne angle, three questions: θ = −1020°

Let θ = −1020°. Answer three questions without a calculator. (a) In which quadrant does the terminal side of θ lie, or does it lie on an axis? (b) Find the exact value of cos(−1020°). (c) Decide whether tan(−1020°) is positive, negative, zero or undefined. Use the odd property of tangent and its period of 180°.

The angle −1020° turns clockwise almost three full turns. It ends on the same terminal side as 60° in Quadrant I.

Part (a) asks for a location. Keep the signed angle −1020° and add full turns: −1020° + 3·360° = −1020° + 1080° = 60°.Adding whole turns of 360° keeps the terminal side the same. Changing the sign to +1020° would give 300°, which is a different place.
60° lies between 0° and 90°, so the terminal side of θ is in Quadrant I. Stop here for part (a).Location depends only on where the terminal point lies after the reduction by full turns.
Part (b): cosine is even, so cos(−1020°) = cos(1020°). There is no outside sign.cos(−t) = cos t for every t.
Remove whole periods of 360°: 1020° − 2·360° = 300°. So cos(1020°) = cos 300°.Cosine has period 360°.
300° is in Quadrant IV, where cosine is positive. Its reference angle is 360° − 300° = 60°. So cos 300° = cos 60° = $1 over 2$ .In Quadrant IV the x-coordinate is positive. The known exact value is cos 60° = $1 over 2$ .
Part (c): tangent is odd, so tan(−1020°) = −tan(1020°). Record the outside negative sign.tan(−t) = −tan t for every t where tangent is defined.
Remove whole periods of 180°: 1020° − 5·180° = 1020° − 900° = 120°. So tan(1020°) = tan 120°.Tangent has period 180°. This reduction is used only for the value, not for the location.
120° is in Quadrant II, where tangent is negative, so tan 120° < 0. Combine this with the outside sign: −(negative) is positive. So tan(−1020°) > 0.For a sign question you multiply the outside sign by the sign of the reduced expression, and then you stop.

Answer

(a) Quadrant I. (b) cos(−1020°) = . (c) tan(−1020°) is positive.

Check −1020° is coterminal with 60°. Then cos 60° = , which matches (b). Also tan 60° = > 0, which matches (c). Both coordinates are positive in Quadrant I, which agrees with (a).

Work to write

−1020° + 1080° = 60°, so the terminal side is in Quadrant I
cos(−1020°) = cos(1020°) = cos(300°) = cos 60° = $1 over 2$
tan(−1020°) = −tan(1020°) = −tan(120°)
tan 120° < 0, so tan(−1020°) > 0

(a) Quadrant I. (b) cos(−1020°) = $1 over 2$ . (c) tan(−1020°) is positive.

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: one extra full turn

Find sin 390°. The input is one extra turn past a known angle.

The two angles stop at the same point.

390° − 360° = 30°.Sine has period 360°.
sin 390° = sin 30° = $1 over 2$ .The 30° sine is a known exact value.

Answer

Check 390° ends in Quadrant I, where sine is positive.

Rung 2Rung 2: an even function with a negative large input

Find cos(−420°). Reverse the input and remove a full turn.

The original input stops to the right of the y-axis.

cos(−420°) = cos 420°.Cosine is even.
420° − 360° = 60°, so cos 420° = cos 60°.Cosine repeats every 360°.
cos 60° = $1 over 2$ .The 60° special triangle gives this coordinate.

Answer

Check The original input reduces to 300°, in Quadrant IV, where cosine is positive.

Rung 3Rung 3: several tangent periods

Find tan $17 π over 4$ . Remove tangent periods to reach a known input.

Equal coordinates produce tangent 1.

4π = $16 π over 4$ , so $17 π over 4$ − 4π = $π over 4$ .Four tangent periods fit in the input.
tan $17 π over 4$ = tan $π over 4$ = 1.Tangent repeats every π, and equal triangle legs give tangent 1.

Answer

Check Removing full turns also gives , whose sine and cosine are equal.

Rung 4Rung 4: a negative reciprocal function

Find csc(− $13 π over 6$ ). Keep the outside sign while reducing.

The negative height has reciprocal −2.

csc(− $13 π over 6$ ) = −csc $13 π over 6$ .Cosecant is odd.
$13 π over 6$ − 2π = $13 π over 6$ − $12 π over 6$ = $π over 6$ .Cosecant has period 2π.
−csc $π over 6$ = −(1 ÷ $1 over 2$ ) = −2.Cosecant is the reciprocal of the known sine.

Answer

−2

Check The original angle reduces to , in Quadrant IV, where cosecant is negative.

Rung 5Rung 5: a zero denominator

Determine sec(−810°). The task is to decide whether an output exists.

The zero x-coordinate blocks secant.

Use even symmetry to consider sec 810°.Secant keeps its output when the input sign changes.
810° − 2 × 360° = 90°, whose point is (0, 1).Remove two full periods.
cos 90° = 0, so secant's denominator is 0.Secant is 1 divided by cosine.
The original secant is undefined.No number times 0 equals 1.

Answer

undefined

Check Directly reducing −810° gives 270°, also with cosine 0.

Rung 6Rung 6: an unfamiliar angle, sign only

Decide the sign of sin(−220°). An exact decimal or special value is not requested.

The original input ends above the x-axis.

sin(−220°) = −sin 220°.Sine is odd.
180° < 220° < 270°, so sin 220° is negative.220° lies in Quadrant III, below the x-axis.
The negative of that negative output is positive.Two sign reversals give a positive result.

Answer

positive

Check Direct reduction gives −220° + 360° = 140°, in Quadrant II, where sine is positive.

Rung 7Rung 7: an unfamiliar fraction of π, sign only

Decide the sign of tan(− $41 π over 8$ ). Find the sign without claiming a special exact value.

The original terminal point gives a negative quotient.

tan(− $41 π over 8$ ) = −tan $41 π over 8$ .Tangent is odd.
5π = $40 π over 8$ , so $41 π over 8$ − 5π = $π over 8$ .Five tangent periods reduce the positive input.
0 < $π over 8$ < $π over 2$ , so tan $π over 8$ is positive.The reduced angle is in Quadrant I.
The outside negative sign makes the original output negative.The sign from the first step must be retained.

Answer

negative

Check A full-turn reduction of the original input is − + 6π = , in Quadrant II, where tangent is negative.

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

✗ Not this: Every angle reduces to a special angle with an elementary exact value.

Removing full periods preserves the leftover angle. It does not turn 17° into 30°.

✓ Instead: Use known exact values where available. For an unfamiliar angle, report the sign or keep an exact expression such as sin 17°.

✗ Not this: sin(−750°) = sin 30° = .

The reduction of the positive input is correct, but the outside negative sign from odd symmetry was lost.

✓ Instead: sin(−750°) = −sin 750° = −sin 30° = −.

Tips and tricks

Write every equality, including any sign outside the function. This makes sign losses visible.
A full-turn reduction is always safe for locating an angle. A π reduction is a value shortcut for tangent and cotangent, and the resulting angle may have a different terminal point.
When an answer is undefined, state which denominator is 0.

Trap. Losing the outside sign while reducing the input changes the expression. Carry it through every equality. Also check axis points before attempting a reciprocal: an undefined value cannot be fixed by moving a negative sign.

Keep in mind

Carry the outside minus through every line: csc(−1110°) = −csc 30° = −2, not 2.
Tangent and cotangent may drop 180° at a time while the other four need 360°: tan 1290° = tan(1290° − 7 × 180°) = tan 30°.
If the angle lands on an axis, read the axis point and check the bottom: 1170° − 1080° = 90°, where x = 0, so sec 1170° is undefined.
Some angles never become special angles: sin 1117° = sin 37°, and the method stops there with the sign, positive, instead of an exact value.

Memory hookMinus, laps, map, value: handle the minus sign, strip whole laps, map the quadrant for the sign, then look up the value. Shrink first, sign second.

Flash cards: say the answer out loud, then flip

Name the four steps in order.

1. The minus sign (even or odd).
2. Remove whole periods.
3. The quadrant gives the sign.
4. The known value.

Sign of cos(−925°)?

Negative: cos(−925°) = cos 925° = cos 205°, and 205° is in Quadrant III.

sec(−780°) = ?

2: secant is even, 780° − 720° = 60°, and sec 60° = 2.

cot 1305° = ?

1: 1305° − 7 × 180° = 45°, and cot 45° = 1.

A student writes csc(−1110°) = 2. What went wrong?

The outside minus from the odd step was dropped. The answer is −2.