What are reduction formulas?

They are the identities that bring the value of a trigonometric function at any angle back to the value at an acute angle. For instance sin 150° = sin 30° and cos 210° = −cos 30°. Thanks to them a table of values from 0° to 90° is all you ever need.

How do I remember the reduction formulas?

Instead of a dozen identities, one rule in two steps. Step one: if the angle is written next to 180° or 360° (that is π or 2π), the function stays the same; next to 90° or 270° (π/2 or 3π/2) it swaps for its cofunction — sine for cosine and back, tangent for cotangent. Step two: take the sign from the quadrant the whole angle falls in, assuming α is acute.

What is sin(180° − α)?

Exactly sin α. The angle 180° − α lies in quadrant II, where the sine is positive, and next to 180° the function does not change. On the unit circle both angles give points with the same y-coordinate, mirrored about the y axis — hence the equality, and hence also the two-solution case of the law of sines.

What is cos(90° + α)?

It equals −sin α. Next to 90° the function swaps for its cofunction (cosine for sine), and 90° + α lies in quadrant II, where the cosine is negative — hence the minus. Check with α = 30°: cos 120° = −1/2 = −sin 30° ✓.

Why is sin(−α) = −sin α while cos(−α) = cos α?

Because a negative angle is measured the other way round, so the point on the unit circle is reflected in the x axis. The reflection flips the sign of the y-coordinate (the sine) and leaves the x-coordinate (the cosine) alone. We say the sine is an odd function and the cosine an even one.

What do the reduction formulas look like in radians?

The same, with 180° written as π and 90° as π/2: sin(π − α) = sin α, cos(π + α) = −cos α, sin(π/2 − α) = cos α, cos(π/2 + α) = −sin α. The rule is identical — next to π and 2π the function stays, next to π/2 and 3π/2 it swaps for its cofunction.

What is a reference angle?

It is the acute angle between the arm of the given angle and the x axis. For 150° it is 30°, for 200° it is 20°, for 320° it is 40°. The magnitude of a function is the same at an angle and at its reference angle; only the sign can differ, and that comes from the quadrant.

Do the reduction formulas work for the tangent?

Yes, and they are even simpler, because the tangent has period 180° rather than 360°: tan(180° + α) = tan α and tan(180° − α) = −tan α. Next to 90° the tangent turns into the cotangent: tan(90° − α) = cot α. The sign always comes from the quadrant, exactly as for the sine and the cosine.

Advanced level

Reduction formulas

Reduction formulas bring the sine, cosine and tangent of any angle back to an acute one. Instead of memorising a dozen identities, see the single rule behind them: at 180° and 360° the function stays, at 90° and 270° it swaps for its cofunction, and the sign comes from the quadrant.

Before you start

This topic builds on earlier ideas. Before you start, it's worth working through the lessons below — they'll make everything click:

The unit circleThe unit circle carries the sine and the cosine from a triangle onto the whole circle. See how the coordinates of a point become the values of the functions, what signs they take in the four quadrants, and what a negative angle or one past a full turn means.
Radian measureA radian is the angle that cuts an arc equal to the radius. See where the 2π of a full turn comes from, how to convert degrees and radians both ways, and why the arc-length and sector-area formulas collapse to a single multiplication in this measure.

Where this is used

Real situations where you count exactly the way this lesson teaches:

Noise-cancelling headphones
A microphone picks up engine noise of amplitude 40 mPa and the speaker plays the same tone shifted by 180°: sin(x + 180°) = −sin x, so the reproduced wave is exactly opposite. The sum 40 · sin x + 40 · sin(x + 180°) = 0 — silence. Active noise cancellation is one reduction formula, executed 48,000 times a second.
Navigation
A ship sails 12 km on a bearing of 215°. The northward component is 12 · cos 215° = −12 · cos 35° = −9.83 km, that is 9.83 km south; the eastward one is 12 · sin 215° = −12 · sin 35° = −6.88 km, that is 6.88 km west. The reduction formula turns an unfamiliar 215° into cos 35° plus two minus signs from quadrant III.
Computer graphics
Rotating a point by 90° counter-clockwise sends (x, y) to (−y, x). That is not a rule to memorise but a pair of reduction formulas: cos(α + 90°) = −sin α and sin(α + 90°) = cos α. The point (3, 4) lands at (−4, 3) — one swap of two numbers instead of two trigonometric calls per pixel.
Mechanical vibration
A mass on a spring has position x(t) = 0.05 · cos(10t) metres. Its velocity runs a quarter period ahead: v(t) = −0.5 · sin(10t) = 0.5 · cos(10t + 90°) m/s. After 0.1 s the position is 0.05 · cos 1 ≈ 0.027 m and the velocity −0.5 · sin 1 ≈ −0.42 m/s — the mass is still on the positive side but already coming back.
Three-phase power
The three phases of an industrial socket are 120° apart: u₂(t) = 325 · sin(ωt − 120°), u₃(t) = 325 · sin(ωt + 120°). At the instant the first phase peaks (ωt = 90°), the second gives 325 · sin(−30°) = −162.5 V and the third 325 · sin 210° = −162.5 V. The three sum to zero at every instant — which is why a balanced load draws no current through the neutral.

All formulas

Supplement of the angle
$\sin(180^\circ - \alpha) = \sin \alpha, \qquad \cos(180^\circ - \alpha) = -\cos \alpha$
quadrant II: sine positive, cosine negative
Half a turn onwards
$\sin(180^\circ + \alpha) = -\sin \alpha, \qquad \cos(180^\circ + \alpha) = -\cos \alpha$
quadrant III: both functions negative
Completing a full turn
$\sin(360^\circ - \alpha) = -\sin \alpha, \qquad \cos(360^\circ - \alpha) = \cos \alpha$
quadrant IV: sine negative, cosine positive
A negative angle
$\sin(-\alpha) = -\sin \alpha, \qquad \cos(-\alpha) = \cos \alpha$
the sine is odd, the cosine even
Complement of the angle
$\sin(90^\circ - \alpha) = \cos \alpha, \qquad \cos(90^\circ - \alpha) = \sin \alpha$
the function swaps for its cofunction
A quarter turn onwards
$\sin(90^\circ + \alpha) = \cos \alpha, \qquad \cos(90^\circ + \alpha) = -\sin \alpha$
the functions swap again, sign from quadrant II
Period of the tangent
$\tan(180^\circ + \alpha) = \tan \alpha$
the tangent repeats every half turn

A table of trigonometric values covers angles from 0° to 90°. There are infinitely many angles. Reduction formulas are the bridge between the two: they bring every value back to an acute angle.

Two separate questions: value and sign

On the unit circle every angle marks a point, and the values of the functions are its coordinates. Reducing to an acute angle therefore asks two separate questions:

What is the magnitude? — that is the reference angle, the acute angle between the arm and the $x$ axis.
What is the sign? — that is the quadrant the arm falls in.

quadrant	range	$\sin$	$\cos$	$\tan$
I	$0^\circ$ – $90^\circ$	$+$	$+$	$+$
II	$90^\circ$ – $180^\circ$	$+$	$-$	$-$
III	$180^\circ$ – $270^\circ$	$-$	$-$	$+$
IV	$270^\circ$ – $360^\circ$	$-$	$+$	$-$

The reference angle is found by subtracting from the nearest horizontal semi-axis:

\text{II: } 180^\circ - \alpha, \qquad \text{III: } \alpha - 180^\circ, \qquad \text{IV: } 360^\circ - \alpha

Find sin 200° and cos 200° from the reference angle.

Next to 180° and 360°: the function stays

Start with the angle $180^\circ - \alpha$ . On the unit circle the two arms sit symmetrically about the $y$ axis.

A reflection in the y axis keeps the y-coordinate (the sine) and flips the sign of the x-coordinate (the cosine).

A reflection in the $y$ axis keeps $y$ and flips the sign of $x$ , so

\sin(180^\circ - \alpha) = \sin \alpha, \qquad \cos(180^\circ - \alpha) = -\cos \alpha

The other two cases read the same way. A rotation by $180^\circ$ sends the point into the diagonally opposite quadrant, so both coordinates change sign:

\sin(180^\circ + \alpha) = -\sin \alpha, \qquad \cos(180^\circ + \alpha) = -\cos \alpha

and a reflection in the $x$ axis (the angle $360^\circ - \alpha$ , which is the same arm as $-\alpha$ ) changes only the $y$ -coordinate:

\sin(360^\circ - \alpha) = -\sin \alpha, \qquad \cos(360^\circ - \alpha) = \cos \alpha

What they share: the function stays the same, and at most the sign changes.

A negative angle: parity

One special case of the last pair deserves its own name, because it turns up everywhere:

\sin(-\alpha) = -\sin \alpha, \qquad \cos(-\alpha) = \cos \alpha

We say the sine is an odd function and the cosine an even one. It shows on the graphs too: the sine curve is symmetric about the origin, the cosine curve about the $y$ axis.

Next to 90° and 270°: the function changes

Here something else happens. The angle $90^\circ - \alpha$ is the complement of $\alpha$ , and on the unit circle its arm is the reflection of $\alpha$ 's arm in the line $y = x$ .

A reflection in the line y = x swaps the coordinates: the x-coordinate becomes the y-coordinate. That is why the cosine turns into the sine.

Reflecting in $y = x$ swaps the coordinates, and since the cosine is the $x$ -coordinate and the sine the $y$ -coordinate, it swaps the two functions as well:

\sin(90^\circ - \alpha) = \cos \alpha, \qquad \cos(90^\circ - \alpha) = \sin \alpha

You have met this one before, in the right triangle: its acute angles add up to $90^\circ$ , and the leg opposite one of them is adjacent to the other.

For $90^\circ + \alpha$ the function swaps in the same way and the sign comes from quadrant II:

\sin(90^\circ + \alpha) = \cos \alpha, \qquad \cos(90^\circ + \alpha) = -\sin \alpha

One rule instead of a table

Every formula in this lesson can be written as two steps. Assume $\alpha$ is acute.

The function. Next to $180^\circ$ and $360^\circ$ (that is $\pi$ and $2\pi$ ) the function stays. Next to $90^\circ$ and $270^\circ$ ( $\frac{\pi}{2}$ and $\frac{3\pi}{2}$ ) it swaps for its cofunction: sine ↔ cosine, tangent ↔ cotangent.
The sign. Find the quadrant the whole angle falls in and take the sign the function on the left-hand side has there.

Reduce cos(270° + α), assuming α is acute.

The same in radians

The notation changes, the rule does not. Substitute $180^\circ = \pi$ and $90^\circ = \frac{\pi}{2}$ :

degrees	radians	$\sin$	$\cos$
$180^\circ - \alpha$	$\pi - \alpha$	$\sin \alpha$	$-\cos \alpha$
$180^\circ + \alpha$	$\pi + \alpha$	$-\sin \alpha$	$-\cos \alpha$
$360^\circ - \alpha$	$2\pi - \alpha$	$-\sin \alpha$	$\cos \alpha$
$90^\circ - \alpha$	$\frac{\pi}{2} - \alpha$	$\cos \alpha$	$\sin \alpha$
$90^\circ + \alpha$	$\frac{\pi}{2} + \alpha$	$\cos \alpha$	$-\sin \alpha$

Given that sin α = 0.6 and α is acute, find sin(π + α) and cos(π/2 + α).

The tangent: shorter period, simpler formulas

The tangent has period $180^\circ$ rather than $360^\circ$ , so a half turn does nothing to it:

\tan(180^\circ + \alpha) = \tan \alpha, \qquad \tan(180^\circ - \alpha) = -\tan \alpha

Next to $90^\circ$ the same swap rule applies — the tangent becomes the cotangent, that is its reciprocal:

\tan(90^\circ - \alpha) = \cot \alpha = \frac{1}{\tan \alpha}

Exercises

For a reference angle type the number of degrees alone (e.g. 20) — it is always between 0° and 90°. For a reduced value the answer is a fraction, sign included (e.g. −3/5).

Practice

Work through a set of exercises — they get harder as you go. At the end you'll see your score and the mistakes worth reviewing.

Exercise 1 of 8Score: 0

the reference angle (°): sin 114°

Common mistakes

A lost minus sign — the magnitude comes from the reference angle, but the sign comes from the quadrant of the whole angle; $\cos 200^\circ$ is negative even though $\cos 20^\circ$ is positive.
Swapping the function next to 180° — next to $180^\circ$ and $360^\circ$ the function stays; it changes only next to $90^\circ$ and $270^\circ$ .
Taking the sign from the function on the right — the sign is decided for the function standing before the reduction. In $\cos(90^\circ + \alpha)$ you look at the cosine in quadrant II, not at the sine.
Measuring the reference angle from the $y$ axis — subtract from $180^\circ$ or $360^\circ$ , never from $90^\circ$ or $270^\circ$ .
Assuming $\alpha$ is acute when it is not — the sign rule assumes $0^\circ < \alpha < 90^\circ$ . Larger angles have to be brought into one turn first.
Confusing the reference angle with the quadrant — $200^\circ$ is quadrant III with reference angle $20^\circ$ ; two different numbers answering two different questions.

Formula card

Topic: Reduction formulas

Supplement of the angle
$\sin(180^\circ - \alpha) = \sin \alpha, \qquad \cos(180^\circ - \alpha) = -\cos \alpha$
quadrant II: sine positive, cosine negative
Half a turn onwards
$\sin(180^\circ + \alpha) = -\sin \alpha, \qquad \cos(180^\circ + \alpha) = -\cos \alpha$
quadrant III: both functions negative
Completing a full turn
$\sin(360^\circ - \alpha) = -\sin \alpha, \qquad \cos(360^\circ - \alpha) = \cos \alpha$
quadrant IV: sine negative, cosine positive
A negative angle
$\sin(-\alpha) = -\sin \alpha, \qquad \cos(-\alpha) = \cos \alpha$
the sine is odd, the cosine even
Complement of the angle
$\sin(90^\circ - \alpha) = \cos \alpha, \qquad \cos(90^\circ - \alpha) = \sin \alpha$
the function swaps for its cofunction
A quarter turn onwards
$\sin(90^\circ + \alpha) = \cos \alpha, \qquad \cos(90^\circ + \alpha) = -\sin \alpha$
the functions swap again, sign from quadrant II
Period of the tangent
$\tan(180^\circ + \alpha) = \tan \alpha$
the tangent repeats every half turn

The angles α and 180° − α sit symmetrically about the y axis. Their y-coordinates are equal (hence sin(180° − α) = sin α) and their x-coordinates opposite (hence cos(180° − α) = −cos α).

The angles α and 90° − α are symmetric about the line y = x. The reflection swaps the coordinates — which is why the sine turns into the cosine.

Frequently asked questions

A luminous unit circle with a radius sweeping out an angle, a sine wave unrolling from it as a streak of light, beside a right triangle drawn in stars, against a cosmic nebula in violet and magenta

Trigonometry

Why sine doesn't depend on the size of the triangle

The height of a tower, the width of a river, the distance to a ship offshore — all of it can be computed without going anywhere near the thing being measured. The whole of trigonometry rests on one observation: the ratio of two sides in a right triangle depends only on the angle, not on how big the triangle is. That is what turns an angle into a number — and lets it be tabulated once and for all.

13 min