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Negative Angle Trigonometric Identities


sin(−θ) = −sin θ
θ40°
Negative angle scenexyOsin θPθP: y = sin θ−sin θP'−θP': y = −sin θy = sin(−θ)
Step 0 of 3

sin θ

0.643

sin(−θ)

-0.643

Derivation
Press Play to step through the proof.
Function
Identity
Parity
Value
Source







Key Terms

• Negative angle identity — a formula relating a trig function evaluated at −θ-\theta to the same function at θ\theta.
• Even function — satisfies f(−x)=f(x)f(-x) = f(x). Its graph is symmetric about the y-axis. cos⁡\cos and sec⁡\sec are even.
• Odd function — satisfies f(−x)=−f(x)f(-x) = -f(x). Its graph is symmetric about the origin. sin⁡\sin, tan⁡\tan, csc⁡\csc, cot⁡\cot are odd.
• Reflection across the x-axis — the geometric operation taking the terminal point P=(cos⁡θ,sin⁡θ)P = (\cos\theta, \sin\theta) to P′=(cos⁡θ,−sin⁡θ)P' = (\cos\theta, -\sin\theta), which is the terminal point of −θ-\theta.
• Parity — whether a function is even or odd, summarized in the Parity column of the formula table.

Switching Between Functions

Six tabs at the top let you select which negative-angle identity to study: sin⁡(−θ)\sin(-\theta), cos⁡(−θ)\cos(-\theta), tan⁡(−θ)\tan(-\theta), csc⁡(−θ)\csc(-\theta), sec⁡(−θ)\sec(-\theta), cot⁡(−θ)\cot(-\theta).

How selection changes the view:
• sin⁡\sin and cos⁡\cos open the geometric proof scene showing the reflection P→P′P \to P' on the unit circle.
• tan⁡\tan, csc⁡\csc, sec⁡\sec, and cot⁡\cot open the derived identity card with the algebraic chain.
• The active tab is highlighted in deep blue.
• The URL updates with ?negFn=...?negFn=... so links you share preserve the selected function.

Clicking any row of the formula table at the bottom also jumps to that function.

Adjusting the Angle θ

Each view exposes a slider for the angle θ\theta in degrees, between 15°15° and 75°75°.

What changes as you slide:
• On geometric scenes, PP moves along the upper unit circle and P′P' follows below as its mirror image.
• The coordinate readouts at PP and P′P' update in real time.
• The verification cards at the bottom recompute both sin⁡θ\sin\theta and sin⁡(−θ)\sin(-\theta) (or cos⁡θ\cos\theta and cos⁡(−θ)\cos(-\theta)) at the new θ\theta.

Sweep the slider to see that sin⁡(−θ)\sin(-\theta) and −sin⁡θ-\sin\theta track together (odd behavior), while cos⁡(−θ)\cos(-\theta) and cos⁡θ\cos\theta stay identical (even behavior).

Playing Through a Geometric Proof

When sin⁡\sin or cos⁡\cos is active, an animated proof unfolds in three steps. A toolbar gives you control:

• Reset — return to step 0 with a blank scene.
• Prev / Next — step one stage at a time.
• Play / Pause — advance automatically.
• Speed selector — 0.5×0.5\times, 1×1\times, 1.5×1.5\times, 2×2\times.

The three steps are: (1) place P at angle θ, (2) reflect across the x-axis to produce P′P' at angle −θ-\theta, (3) read off the identity from the coordinates of P′P'.

Reading the Geometric Scene

The SVG shows the unit circle with two terminal points:

• P at angle θ\theta above the x-axis, with coordinates (cos⁡θ,sin⁡θ)(\cos\theta, \sin\theta).
• P' at angle −θ-\theta below the x-axis, with coordinates (cos⁡θ,−sin⁡θ)(\cos\theta, -\sin\theta).

Reflection across the x-axis is the key operation:
• Preserves the x-coordinate — so cosine comes out even, cos⁡(−θ)=cos⁡θ\cos(-\theta) = \cos\theta.
• Flips the sign of the y-coordinate — so sine comes out odd, sin⁡(−θ)=−sin⁡θ\sin(-\theta) = -\sin\theta.

A comparison overlay highlights the shared horizontal foot and the equal-magnitude, opposite-sign vertical legs.

Working with Derived Identities

Selecting tan⁡(−θ)\tan(-\theta), csc⁡(−θ)\csc(-\theta), sec⁡(−θ)\sec(-\theta), or cot⁡(−θ)\cot(-\theta) opens a different card layout. Instead of a unit-circle picture, it shows the algebraic derivation as a chain of equations.

Layout of the derived card:
• A short intro explains which earlier identity the current one rests on.
• Jump buttons link directly to the geometric proofs of sin⁡(−θ)\sin(-\theta) or cos⁡(−θ)\cos(-\theta).
• A multi-line derivation block shows each manipulation with a brief side note.
• Verification cards confirm both sides match numerically.

Two derived identities preserve sign (sec⁡\sec, like cos⁡\cos, is even); four flip sign (tan⁡\tan, csc⁡\csc, cot⁡\cot, like sin⁡\sin, are odd).

Reading the Formula Table

A reference table beneath every scene lists all six negative-angle identities at once:

• Function column — the trig function with −θ-\theta argument.
• Identity column — the right-hand side.
• Parity column — labels each as even or odd.
• Value column — the numeric value at the current θ\theta.
• Source column — labels each as geometric (sin⁡\sin, cos⁡\cos) or via X for the derived forms.

Click any row to make that function active. The current row gets a deep-blue left border and tinted background.

Verifying Identities Numerically

Every scene includes two metric cards that compute the function at +θ+\theta and at −θ-\theta for the active row.

Example for sin⁡\sin:
• Left card shows sin⁡θ\sin\theta.
• Right card shows sin⁡(−θ)\sin(-\theta).

For odd functions, the two values are equal in magnitude and opposite in sign. For even functions, they are identical. Sweeping the slider while watching the cards is a fast empirical check across all θ\theta, and the formula table mirrors this across all six functions at once.

Geometric Proofs: sin(-θ) and cos(-θ)

The two foundational identities come from reflecting the terminal point across the x-axis.

sin(-θ) = -sin θ — reflection flips the y-coordinate:
P=(cos⁡θ,sin⁡θ)  →  P′=(cos⁡θ,−sin⁡θ)P = (\cos\theta, \sin\theta) \;\to\; P' = (\cos\theta, -\sin\theta)

Since the y-coordinate of P′P' is sin⁡(−θ)\sin(-\theta) by definition, sin⁡(−θ)=−sin⁡θ\sin(-\theta) = -\sin\theta. Sine is odd.

cos(-θ) = cos θ — reflection preserves the x-coordinate. The x-coordinate of P′P' equals the x-coordinate of PP, which is cos⁡θ\cos\theta. Therefore cos⁡(−θ)=cos⁡θ\cos(-\theta) = \cos\theta. Cosine is even.

Each is treated stage by stage below: the sine form and the cosine form.

For full coverage with proofs in all quadrants, see the negative angle identities theory page.

Derived Identities: tan, csc, sec, cot

The four remaining identities follow from the two geometric ones:

tan(-θ) = -tan θ (odd) — tan⁡=sin⁡/cos⁡\tan = \sin/\cos:
tan⁡(−θ)=sin⁡(−θ)cos⁡(−θ)=−sin⁡θcos⁡θ=−tan⁡θ\tan(-\theta) = \frac{\sin(-\theta)}{\cos(-\theta)} = \frac{-\sin\theta}{\cos\theta} = -\tan\theta


csc(-θ) = -csc θ (odd) — reciprocal of an odd function is odd: csc⁡(−θ)=1/sin⁡(−θ)=−1/sin⁡θ=−csc⁡θ\csc(-\theta) = 1/\sin(-\theta) = -1/\sin\theta = -\csc\theta.

sec(-θ) = sec θ (even) — reciprocal of an even function is even: sec⁡(−θ)=1/cos⁡(−θ)=1/cos⁡θ=sec⁡θ\sec(-\theta) = 1/\cos(-\theta) = 1/\cos\theta = \sec\theta.

cot(-θ) = -cot θ (odd) — reciprocal of tangent: cot⁡(−θ)=1/tan⁡(−θ)=−1/tan⁡θ=−cot⁡θ\cot(-\theta) = 1/\tan(-\theta) = -1/\tan\theta = -\cot\theta.

For full derivations, see the trigonometric identities page and the reciprocal identities page.

Why Negative Angle Identities Matter

These identities reveal the symmetry structure of trigonometric functions:

• Graph symmetry — the parity rules predict whether each graph is symmetric about the y-axis (even) or about the origin (odd) without plotting points.
• Simplification — replace any f(−θ)f(-\theta) with ±f(θ)\pm f(\theta) instantly, halving the cases to consider.
• Fourier series — even functions expand into cosines only, odd functions into sines only.
• Integration — odd integrands over symmetric intervals like [−a,a][-a, a] integrate to zero.
• Solving equations — paired solutions θ\theta and −θ-\theta are predictable from parity.

All five uses come down to one of two facts, each proved in a single picture: sine flips sign, cosine does not.

For applications and examples, see the trigonometric identities applications page.

The Sine Negative-Angle Identity

The identity sin⁡(−θ)=−sin⁡θ\sin(-\theta) = -\sin\theta is proved by reflecting the terminal point across the x-axis and reading its new y-coordinate.
xyOsin θPθP: y = sin θ−sin θP′−θP′: y = −sin θy = sin(−θ)sin(−θ) = −sin θ · −0.643 = −0.643
The complete sine proof, frozen

P and its mirror image P′, with the two amber legs of equal length pointing opposite ways — the whole content of sin(−θ) = −sin θ.

Its three stages: place the point, mirror it, and read off the y-coordinate.

Sine is the odd function of the pair, and three of the four derived identities inherit their sign flip from it: cosecant directly, tangent through the quotient, and cotangent through tangent.

The Cosine Negative-Angle Identity

The identity cos⁡(−θ)=cos⁡θ\cos(-\theta) = \cos\theta comes from the same reflection, read along the other axis: mirroring across the x-axis cannot change an x-coordinate.
xyOcos θPθP: x = cos θP′−θP′: x = cos θx = cos(−θ)cos(−θ) = cos θ · 0.766 = 0.766
The complete cosine proof, frozen

One indigo leg serving both points: mirroring across the x-axis cannot move a foot that sits on the x-axis.

Its three stages: place the point, mirror it, and read off the x-coordinate.

Under the cosine tab the tool hides the vertical legs entirely and draws a faint connector between PP and P′P' instead — the picture is arguing that the two points sit on one vertical line, which is exactly what "same x-coordinate" means. Cosine is the even function of the pair, and secant is the only derived identity that inherits that evenness.

The Tangent Negative-Angle Identity

Tangent is a quotient of one odd function and one even function, so exactly one sign flips and the quotient comes out odd.
tan(−θ) = −tan θtan(−θ)=sin(−θ) / cos(−θ)definition=(−sin θ) / (cos θ)sin(−θ) = −sin θ, cos(−θ) = cos θ=−tan θsimplifyat θ = 40° : tan(−θ) = −0.839 = −tan θ
tan(−θ), derived

Odd over even: the numerator flips, the denominator does not, so the quotient flips.

tan⁡(−θ)=sin⁡(−θ)cos⁡(−θ)=−sin⁡θcos⁡θ=−tan⁡θ\tan(-\theta) = \frac{\sin(-\theta)}{\cos(-\theta)} = \frac{-\sin\theta}{\cos\theta} = -\tan\theta


This is the general rule in miniature: odd divided by even is odd. Both ingredients are proved geometrically — the sine identity supplies the minus sign, the cosine identity supplies the unchanged denominator — and the card's jump buttons lead to each. Cotangent then inherits the flip from tangent.

The Cosecant Negative-Angle Identity

Cosecant is the reciprocal of sine, and a reciprocal keeps the parity of what it inverts.
csc(−θ) = −csc θcsc(−θ)=1 / sin(−θ)definition=1 / (−sin θ)sin(−θ) = −sin θ=−csc θsimplifyat θ = 40° : csc(−θ) = −1.556 = −csc θ
csc(−θ), derived

The minus sign travels straight out of the denominator — a reciprocal keeps its parity.

csc⁡(−θ)=1sin⁡(−θ)=1−sin⁡θ=−csc⁡θ\csc(-\theta) = \frac{1}{\sin(-\theta)} = \frac{1}{-\sin\theta} = -\csc\theta


The minus sign moves out of the denominator untouched, so cosecant is odd for the same reason sine is. Note what the identity does not fix: cosecant is undefined wherever sin⁡θ=0\sin\theta = 0, and negating the angle does not rescue it — if one side is undefined, so is the other. The slider's 15°15°–75°75° range stays clear of those angles.

The Secant Negative-Angle Identity

Secant is the reciprocal of cosine, so it is the one derived identity with no sign change at all.
sec(−θ) = sec θsec(−θ)=1 / cos(−θ)definition=1 / cos θcos(−θ) = cos θ=sec θsimplifyat θ = 40° : sec(−θ) = 1.305 = sec θ
sec(−θ), derived

The one derived card with no sign anywhere in it, because cosine had none to give.

sec⁡(−θ)=1cos⁡(−θ)=1cos⁡θ=sec⁡θ\sec(-\theta) = \frac{1}{\cos(-\theta)} = \frac{1}{\cos\theta} = \sec\theta


Nothing flips because nothing flipped in the cosine identity it rests on. Secant and cosine are the two even functions in the table; the other four are odd. Sweeping the slider is the quickest way to see it — the two verification cards for secant never separate, while the cards for every odd function stay opposite in sign.

The Cotangent Negative-Angle Identity

Cotangent is the reciprocal of tangent, so it flips sign for the same reason tangent does — one step further removed from the geometry.
cot(−θ) = −cot θcot(−θ)=1 / tan(−θ)definition=1 / (−tan θ)tan(−θ) = −tan θ=−cot θsimplifyat θ = 40° : cot(−θ) = −1.192 = −cot θ
cot(−θ), derived

Two steps from the picture: cotangent inverts tangent, which already inherited the flip from sine.

cot⁡(−θ)=1tan⁡(−θ)=1−tan⁡θ=−cot⁡θ\cot(-\theta) = \frac{1}{\tan(-\theta)} = \frac{1}{-\tan\theta} = -\cot\theta


Cotangent sits two derivations from the picture: it depends on tangent, which depends on sine and cosine. It can also be read straight off the definition cot⁡θ=cos⁡θ/sin⁡θ\cot\theta = \cos\theta / \sin\theta — even over odd, which is odd — and the two routes agree, as they must.

Sine Proof, Step 1: Place P at Angle θ

The sine proof opens with a single point: PP on the unit circle at angle θ\theta above the x-axis, with the vertical leg from the axis up to PP drawn in amber.
xyOsin θPθP: y = sin θ
Step 1: place P at angle θ

One point, one amber leg. On a unit circle that leg’s signed length is the y-coordinate itself.

That leg has signed length sin⁡θ\sin\theta, and because the circle has radius 11 the length is the y-coordinate of PP — no scaling in between. The red arc at the origin marks the angle θ\theta measured counter-clockwise, the positive direction.

Under this tab the tool runs in its sine-only mode, hiding the horizontal cos⁡θ\cos\theta leg so that nothing competes with the quantity being tracked. The cosine proof hides the opposite one.

Sine Proof, Step 2: Mirror P Across the x-Axis

Reflecting PP across the x-axis produces P′P', sitting directly below at angle −θ-\theta — the same rotation measured clockwise.
xyOsin θPθP: y = sin θ−sin θP′−θP′: y = −sin θ
Step 2: mirror across the x-axis

P′ appears directly below, same distance from the axis, opposite side — equal magnitude, flipped sign.

The reflection is what carries the whole argument, and it does two things at once: it leaves the x-coordinate exactly where it was, and it flips the sign of the y-coordinate. The second amber leg has the same length as the first but points the other way, so its signed length is −sin⁡θ-\sin\theta.

Nothing has been proved yet — so far this is only a construction. The claim arrives when the picture is read as a statement about the angle −θ-\theta, which is the next step.

Sine Proof, Step 3: Read Off sin(-θ)

P′P' is the terminal point of the angle −θ-\theta on the unit circle, so by the definition of sine its y-coordinate is sin⁡(−θ)\sin(-\theta).
xyOsin θPθP: y = sin θ−sin θP′−θP′: y = −sin θy = sin(−θ)sin(−θ) = −sin θ · −0.643 = −0.643
Step 3: read off sin(−θ)

The label y = sin(−θ) names the coordinate the picture already gave as −sin θ.

sin⁡(−θ)=−sin⁡θ\sin(-\theta) = -\sin\theta


The proof is a matter of naming the same number twice. The picture gives that y-coordinate as −sin⁡θ-\sin\theta; the definition gives it as sin⁡(−θ)\sin(-\theta); therefore the two are equal. At the tool's opening angle of 40°40° both verification cards read −0.643-0.643.

That single sign flip is the definition of an odd function, and it propagates through tangent, cosecant and cotangent.

Cosine Proof, Step 1: Place P at Angle θ

The cosine proof begins from the same point PP at angle θ\theta, but with the indigo horizontal leg drawn instead of the vertical one.
xyOcos θPθP: x = cos θ
Step 1: place P at angle θ

Same point, indigo leg instead: the horizontal distance from the y-axis, equal to cos θ.

That leg runs from the origin to the foot of PP on the x-axis, and its length is cos⁡θ\cos\theta — again equal to the coordinate itself, because the radius is 11.

The tool switches to its cosine-only mode here, hiding the vertical legs and the right-angle markers that the sine proof relies on. Same circle, same point, different quantity in view.

Cosine Proof, Step 2: Mirror P Across the x-Axis

The same reflection places P′P' at angle −θ-\theta, and a faint vertical connector is drawn between PP and P′P'.
xyOcos θPθP: x = cos θP′−θP′: x = cos θ
Step 2: mirror across the x-axis

The faint connector is the argument — a vertical segment means one shared x-coordinate.

That connector is the argument. Two points joined by a vertical segment have the same x-coordinate by definition, and the horizontal indigo leg — drawn once — serves both of them. The reflection moved the point without moving its foot.

Compare with the sine version of this step: the same construction supports both proofs, and the tool draws whichever consequence the active tab is about.

Cosine Proof, Step 3: Read Off cos(-θ)

P′P' is the terminal point of −θ-\theta, so its x-coordinate is cos⁡(−θ)\cos(-\theta) — and that x-coordinate is the one PP already had.
xyOcos θPθP: x = cos θP′−θP′: x = cos θx = cos(−θ)cos(−θ) = cos θ · 0.766 = 0.766
Step 3: read off cos(−θ)

x = cos(−θ) labels the same foot that was already labelled cos θ. No sign appears.

cos⁡(−θ)=cos⁡θ\cos(-\theta) = \cos\theta


No sign appears anywhere, which is the whole content of the result: cosine is even. At 40°40° both verification cards read 0.7660.766, and they stay locked together for every angle the slider reaches.

Only secant inherits this evenness; the remaining three derived identities take their sign flip from sine instead.