• Pythagorean identity — a trigonometric identity derived from sin2θ+cos2θ=1, itself a consequence of Pythagoras' theorem on a right triangle with hypotenuse 1. • Base identity — sin2θ+cos2θ=1, the geometric foundation for all six function forms in this tool. • Derived identity — one obtained by dividing the base identity by sin2θ or cos2θ, then rearranging. • Reciprocal function — csc=1/sin, sec=1/cos, cot=1/tan. • Positive root — when solving x2=y for x, the explorer assumes θ is in the first quadrant so all six functions are positive.
Switching Between Functions
Six tabs at the top let you pick which Pythagorean form to study: sinθ, cosθ, tanθ, cscθ, secθ, cotθ.
How selection changes the view: • sin and cos open the geometric proof scene with a step-by-step animation built on a right triangle inside the unit circle. • tan, csc, sec, and cot open the derived identity card with the algebraic chain. • The active tab is highlighted in deep blue. • The URL updates with ?fn=... so links you share preserve the selected function.
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 angleθ in degrees, between 10° and 80° (first quadrant).
What changes as you slide: • On geometric scenes, the triangle inside the unit circle reshapes in real time. • The number readout shows the exact degree value. • The verification cards at the bottom recompute both sides of the identity at the new θ.
When sin or cos is active, an animated proof unfolds in six steps. A toolbar gives you control:
• Reset — return to step 0 with a blank scene. • Prev / Next — step through one stage at a time. • Play / Pause — advance automatically. • Speed selector — 0.5×, 1×, 1.5×, 2×.
Each step adds one element (radii, triangle fill, bisector, half-angles, leg labels sinθ and cosθ, final metrics). The right panel logs each step with reasoning, and every stage has its own write-up below, beginning with the setup.
Reading the Geometric Scene
The SVG shows the unit circle with two radii OA and OB meeting at the center O, with a perpendicular bisector OM.
Elements that appear across the steps: • Indigo chordAB — the base of the isosceles triangle. • Blue segmentOM — the bisector, equal to cosθ in right triangle OMA. • Half-chord labels — sinθ on segment MA. • Right-angle mark at M — the key to applying Pythagoras.
Once the legs are labeled, the identity sin2θ+cos2θ=1 follows from leg² + leg² = hypotenuse² with hypotenuse 1.
Working with Derived Identities
Selecting tanθ, cscθ, secθ, or cotθ opens a different card layout. Instead of a triangle, it shows the algebraic derivation as a chain of equations.
Layout of the derived card: • A short intro explains which manipulation (divide by sin2θ or cos2θ) produces the identity. • Jump buttons link directly to the geometric proofs of the source identitiessinθ and cosθ. • A multi-line derivation block shows each manipulation with a side note. • Verification cards confirm both sides match numerically.
This split keeps the geometric idea isolated to the base identity and treats the other four as algebraic consequences.
Reading the Formula Table
A reference table beneath every scene lists all six Pythagorean identities at once:
• Function column — the active trig function. • Identity column — the identity expressed as a square root. • Value column — the numeric value at the current θ. • Source column — labels each as geometric (sinθ, cosθ) or via sin, cos 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 both sides of the active identity at the current θ.
Example for sinθ: • Left card shows sinθ. • Right card shows 1−cos2θ.
The two numbers always match (within rounding to three decimals). Sweeping the slider while watching the cards is a fast empirical check that the identity holds for every θ in the first quadrant. The formula table mirrors this across all six functions simultaneously.
Geometric Proof: sin²θ + cos²θ = 1
The base identity is proved directly from a right triangle inscribed in the unit circle.
In right triangle OMA: • Hypotenuse OA=1 (a radius of the unit circle). • Leg OM=cosθ. • Leg MA=sinθ.
Applying Pythagoras:
sin2θ+cos2θ=1
Solving for sinθ or cosθ and taking the positive root gives the two geometric identities:
For step-by-step derivations and the unsigned forms valid in all quadrants, see the trigonometric identities page and the reciprocal identities page.
Why Pythagorean Identities Matter
The three Pythagorean identities are the most-used identities in trigonometry:
• Simplification — convert expressions in sin2 to cos2 form and vice versa. • Integration — u-substitution in integrals like ∫sec2θdθ=tanθ+C relies on 1+tan2θ=sec2θ. • Equation solving — quadratic equations in sinθ or cosθ frequently emerge after substitution. • Proofs of other identities — sum, difference, double-angle, and half-angle identities all use Pythagorean relations along the way.
Every one of these leans on a single line of algebra, the base identity, or on one of the five forms rearranged from it.
For applications and worked examples, see the trigonometric identities applications page.
Related Concepts and Tools
Continue exploring with these connected resources:
• Double Angle Identities — formulas for sin(2θ), cos(2θ), tan(2θ) built on Pythagoras. • Half Angle Identities — formulas for sin(α/2) and friends, derived using sin2+cos2=1. • Sum and Difference Identities — additive companions to Pythagoras. • Unit Circle — geometric setup for every identity in this tool. • Trigonometric Functions Graphs — see how sin, cos, tan and their reciprocals evolve as θ varies. • Triangle Explorer — interactive triangles with built-in Pythagoras verification. • Negative Angle Identities — parity does not disturb sin2+cos2=1, and this tool shows why. • Supplementary Angle Identities — reflection across the y-axis leaves every Pythagorean relation intact. • Basic Trigonometric Identities — the reciprocal and quotient identities that turn one Pythagorean identity into three.
The Sine Pythagorean Identity
The identity sinθ=1−cos2θ is proved geometrically in the explorer, by reading the legs of a right triangle whose hypotenuse is a radius of the unit circle.
The complete sine proof, frozen
The finished right triangle OMA: hypotenuse 1, legs sin θ and cos θ, with the identity solved for the half-chord.
Everything else on the page rests on this one. The cosine form is the same equation solved for the other leg, and all four derived identities begin by dividing sin2θ+cos2θ=1 through by one of its two terms.
The Cosine Pythagorean Identity
The identity cosθ=1−sin2θ comes out of the same right triangle, solved for the other leg.
The complete cosine proof, frozen
The same triangle, read for the other leg — the bisector OM, whose length is cos θ.
Because sin2θ+cos2θ=1 treats its two terms alike, this proof and the sine proof are the same argument until the fifth step, where one subtracts cos2θ and the other subtracts sin2θ. Switching tabs between them leaves the picture unchanged and rewrites only the last two entries of the step log.
The Tangent Pythagorean Identity
Tangent needs no new geometry. Dividing the base identity by cos2θ converts it into a statement about tangent and secant.
tan θ, derived
Five algebraic lines: divide the base identity by cos²θ, recognise tan and sec, solve, take the root.
cos2θsin2θ+cos2θ=cos2θ1⟹tan2θ+1=sec2θ
Solving that for the squared tangent and taking the root gives the form the tool displays:
tanθ=sec2θ−1
The same three lines, stopped one step earlier and solved the other way, give the secant form — which is why both cards show the identical derivation down to their fourth line. The division is only legal where cosθ=0; inside the slider's first-quadrant range that never bites, but at θ=90° both tangent and secant blow up together.
The Cosecant Pythagorean Identity
Cosecant comes from the other division: dividing the base identity by sin2θ brings cotangent and cosecant into play.
csc θ, derived
The divide-by-sin²θ chain, solved for the cosecant rather than the cotangent.
sin2θsin2θ+cos2θ=sin2θ1⟹1+cot2θ=csc2θ
Solving for the squared cosecant leaves the identity in the tool's square-root form:
cscθ=1+cot2θ
No root is ever taken of a negative number here: cot2θ cannot be negative, so this form is safe for every θ where cotangent is defined. The cotangent form shares its first three lines, and the division that produced both is undefined only where sinθ=0.
The Secant Pythagorean Identity
Secant is the second harvest of the divide-by-cos2θ derivation: the same relation tan2θ+1=sec2θ, solved for the other unknown.
sec θ, derived
The same first three lines as the tangent card, finished the other way round.
sec2θ=1+tan2θ⟹secθ=1+tan2θ
This is the form that makes trigonometric substitution work in calculus: any expression of the shape 1+x2 becomes a single secant after the substitution x=tanθ. Its partner in the same derivation is the tangent form, and both descend from the base identity proved in the geometric scene.
The Cotangent Pythagorean Identity
Cotangent is the second harvest of the divide-by-sin2θ derivation, taken from 1+cot2θ=csc2θ.
cot θ, derived
Subtracting inside the root instead of adding — the cotangent counterpart of the tangent form.
cot2θ=csc2θ−1⟹cotθ=csc2θ−1
Unlike its partner the cosecant form, this one subtracts inside the root, so it needs csc2θ≥1 — true for every angle, since ∣sinθ∣≤1. The pairing mirrors what happens on the cosine side of the family: tangent subtracts inside its root for the same reason.
Sine Proof, Step 1: Setup
The sine proof opens with two radii OA and OB of length 1 drawn from the center O of the unit circle, joined by the chord AB.
Step 1: the unit setup
Two radii of length 1 from O, closed by the chord AB. No claim yet — only the fixed hypotenuse.
Nothing has been claimed yet. The only fact on the table is the one that carries the whole argument: both radii are exactly 1 long, so any right triangle built inside this figure inherits a hypotenuse of 1.
The figure is isosceles by construction, which is what makes the next move — dropping the perpendicular — split it into two matching halves.
Sine Proof, Step 2: Bisect
Dropping OM perpendicular to the chord AB produces right triangle OMA, with the right angle at M.
Step 2: bisect
OM lands perpendicular on AB, splitting the figure into two congruent right triangles.
Because triangle OAB is isosceles, that perpendicular also bisects the angle at O: the two arcs marked at the center are equal, and each one is the θ the slider controls. The scene draws them only above 22°, so at very small angles the marks disappear while the geometry stays the same.
The small square at M is the reason for this step. Pythagoras applies to right triangles and nothing else, so the proof cannot proceed until one exists.
Sine Proof, Step 3: Identify the Legs
In right triangle OMA the hypotenuse is OA=1 and the angle at O is θ, which makes the two legs the basic ratios themselves.
Step 3: the legs named
A hypotenuse of 1 makes the legs the ratios themselves: MA = sin θ, OM = cos θ.
MA=sinθOM=cosθ
With a hypotenuse of 1, "opposite over hypotenuse" collapses to "opposite". That is why the unit circle is the natural home for these identities: the ratios stop being ratios and become plain lengths you can measure off the picture.
Both labels appear at once, because the same triangle carries both. The cosine proof reads the very same figure — it simply keeps its eye on OM instead of MA.
Sine Proof, Step 4: Pythagoras
Applying the Pythagorean theorem to triangle OMA — leg squared plus leg squared equals hypotenuse squared — turns the picture into the base identity.
Step 4: Pythagoras
Leg squared plus leg squared equals 1 — the base identity, read straight off the triangle.
sin2θ+cos2θ=1
This one line is the source of every result on the page. The cosine identity is it, rearranged; the four derived identities are it, divided through.
Note what the theorem is being applied to: not an abstract triangle, but one whose hypotenuse was fixed at 1 back in the setup. That is where the 1 on the right-hand side comes from.
Sine Proof, Step 5: Solve for sin²θ
Subtracting cos2θ from both sides isolates the squared sine.
Step 5: solve for sin²θ
The banner carries the algebra; the picture is unchanged, because nothing new was constructed.
sin2θ=1−cos2θ
This is where the two geometric proofs part company. The cosine proof subtracts the other term at exactly this point; up to here the two are the same proof with the same picture, which is why switching tabs mid-animation changes nothing on screen.
Sine Proof, Step 6: Take the Positive Root
Taking the square root finishes the identity, and the first-quadrant restriction decides the sign.
Step 6: the positive root
The finished identity, with its value at the frozen angle of 35°.
sinθ=1−cos2θ
The slider's 10°–80° range keeps θ in the first quadrant, where sinθ>0, so the positive root is the correct one. Outside that range the squared statement from the previous step still holds, but the sign in front of the root follows the quadrant.
At the tool's opening angle of 35° the two verification cards both read 0.574 — the same number reached along two independent routes.
Cosine Proof, Step 1: Setup
The cosine proof starts from precisely the figure the sine proof starts from: two unit radii from O, closed off by the chord AB.
Step 1: same setup, other leg
Identical construction to the sine proof — the target, not the figure, is what differs.
The construction is identical because the target is a rearrangement of the same equation. What differs is which leg the proof is chasing — here it is OM, the bisector, rather than the half-chord.
Because the two proofs share a figure, the tool renders the same scene under both tabs and changes only the wording in the step log.
Cosine Proof, Step 2: Bisect
The same perpendicular OM is dropped onto AB, producing the same right triangle OMA with its right angle at M.
Step 2: bisect
The indigo bisector is now the quantity of interest, not just a construction line.
The bisector is drawn in indigo and now carries extra weight: it is not just a construction line but the quantity the proof is about. Its length will turn out to be cosθ exactly.
Everything said about the bisection in the sine version of this step applies unchanged — equal half-angles at O, two congruent right triangles, one right angle to work with.
Cosine Proof, Step 3: Identify the Legs
The hypotenuse is still OA=1 and the angle at O is still θ, so the legs read off the same way — but this time the interest is in the adjacent one.
Step 3: the adjacent leg
OM sits adjacent to θ, so against a unit hypotenuse it is exactly cos θ.
OM=cosθMA=sinθ
The bisector OM sits adjacent to the angle θ, so with a unit hypotenuse it is the cosine. Reading the same figure for the opposite leg is what the sine proof does at this step.
Cosine Proof, Step 4: Pythagoras
Pythagoras applied to triangle OMA produces the base identity again — the two proofs reach the same equation from the same picture.
Step 4: Pythagoras
The same theorem on the same triangle yields the same equation as the sine proof.
sin2θ+cos2θ=1
There is genuinely only one theorem here, appearing twice under two tabs. What makes the identities different is the next step, not this one: the sine proof will isolate one term and this proof will isolate the other.
Cosine Proof, Step 5: Solve for cos²θ
Subtracting sin2θ from both sides isolates the squared cosine.
Step 5: solve for cos²θ
The other term is subtracted this time — the single point where the two proofs diverge.
cos2θ=1−sin2θ
Compare with the sine proof's fifth step: identical algebra, opposite term removed. The symmetry of sin2θ+cos2θ=1 is what makes both readings equally valid, and it is the reason the tool can present the two as separate proofs sharing one figure.
Cosine Proof, Step 6: Take the Positive Root
The square root closes the argument, with the first quadrant again fixing the sign.
Step 6: the positive root
The cosine identity, checked at 35° where both sides read 0.819.
cosθ=1−sin2θ
Cosine is positive throughout the slider's 10°–80° range, so the positive root holds there. In the second and third quadrants cosine turns negative and the root would need a minus sign in front — the equation for cos2θ from the previous step survives everywhere, but its square root does not.