The Pythagorean identity says that sin2θ + cos2θ = 1 for every angle θ. It is the Pythagorean theorem in trigonometric form: on a circle of radius 1, the point at angle θ has coordinates (cosθ, sinθ), and those coordinates are the legs of a right triangle whose hypotenuse is the radius, 1.
An identity is an equation that is true for every value of the variable, not just for one solution. This one has two companions, 1 + tan2θ = sec2θ and 1 + cot2θ = csc2θ, and together they are called the Pythagorean identities.
Deriving sin²θ + cos²θ = 1 from the unit circle
The unit circle is the circle of radius 1 centered at the origin. Measure an angle θ counterclockwise from the positive x-axis and mark the point P where its arm meets the circle. In trigonometry, the coordinates of P define the cosine and sine:
P = (cosθ, sinθ)Now drop a vertical line from P to the x-axis. You get a right triangle with a horizontal leg of length |cosθ|, a vertical leg of length |sinθ|, and a hypotenuse equal to the radius, 1. The Pythagorean theorem gives
|cosθ|2 + |sinθ|2= 12cos2θ + sin2θ= 1Squaring removes the absolute value signs, because a negative number squared is positive. That is why the identity survives in every quadrant, even though sine and cosine change sign.
Two details are worth knowing. First, sin2θ is shorthand for (sinθ)2, the sine squared, not the sine of θ2. Second, when P sits on an axis (at 0°, 90°, 180° or 270°) the triangle flattens out, but the identity still holds directly: at 90°, P = (0, 1) and 02 + 12 = 1.
You can say the same thing without triangles. The point (x, y) lies on the unit circle exactly when x2 + y2 = 1, which is the equation of a circle with radius 1. Substituting x = cosθ and y = sinθ gives the identity again.
The same fact in any right triangle
For an acute angle θ in a right triangle, SOH CAH TOA defines the ratios: if a is the leg opposite θ, b the leg next to it and c the hypotenuse, then sinθ = a over c and cosθ = b over c. Add their squares:
sin2θ + cos2θ = a2 over c2 + b2 over c2 = a2 + b2 over c2 = c2 over c2 = 1The step a2 + b2 = c2 is the theorem itself. Read backwards, multiplying the identity by c2 gives a2 + b2 = c2. The two statements are the same fact, one about lengths and one about ratios.
Key ideasin2θ + cos2θ = 1 is a2 + b2 = c2 divided by c2. Any right triangle, scaled so its hypotenuse is 1, has legs sinθ and cosθ.
Try it on the 3-4-5 triangle. For the angle opposite the side of 3, sinθ = 3 over 5 and cosθ = 4 over 5, and 9 over 25 + 16 over 25 = 25 over 25 = 1. The exact values for 30°, 45° and 60° from the 30-60-90 and 45-45-90 triangles pass the same test.
Because the identity is built on the theorem, using it to prove the Pythagorean theorem would be circular reasoning. For a long time that made "trigonometric proofs" of the theorem look impossible; the page on trigonometric proofs explains how mathematicians have found ways around the problem.
The three Pythagorean identities
The other two identities come from dividing the first one by cos2θ or by sin2θ. They use the reciprocal functions secθ = 1 over cosθ, cscθ = 1 over sinθ and cotθ = cosθ over sinθ.
sin2θ over cos2θ + cos2θ over cos2θ = 1 over cos2θ⇒ tan2θ + 1 = sec2θsin2θ over sin2θ + cos2θ over sin2θ = 1 over sin2θ⇒ 1 + cot2θ = csc2θThe first holds whenever cosθ ≠ 0, and the second whenever sinθ ≠ 0; at the excluded angles the functions involved are undefined. Each one is again the Pythagorean theorem, this time in a right triangle with a leg of 1:
| Identity | Rearranged forms | Valid when |
|---|---|---|
| sin2θ + cos2θ = 1 | sin2θ = 1 − cos2θ, cos2θ = 1 − sin2θ | every angle |
| 1 + tan2θ = sec2θ | tan2θ = sec2θ − 1, sec2θ − tan2θ = 1 | cosθ ≠ 0 |
| 1 + cot2θ = csc2θ | cot2θ = csc2θ − 1, csc2θ − cot2θ = 1 | sinθ ≠ 0 |
Memorize the first identity and you never need to memorize the other two: divide by cos2θ or sin2θ whenever you need them.
Finding cos θ from sin θ, with the right sign
The most common use of the identity is to find one trig value from another. Solving for cosine gives
cosθ = ±√1 − sin2θThe identity gives the size; the quadrant gives the sign. Cosine is the x-coordinate of P and sine is the y-coordinate, so each is positive or negative depending on where P lies.
- Use the identity to get the size. Substitute the value you know and take the square root.
- Use the quadrant to choose the sign. Ask whether the x-coordinate (cosine) or y-coordinate (sine) is positive there.
- Find the rest by division. tanθ = sinθ over cosθ, and the reciprocals follow.
Example 1: sine given, Quadrant II
sinθ = 3 over 5 and θ is in Quadrant II. Find cosθ and tanθ.
cos2θ= 1 − (3 over 5)2 = 1 − 9 over 25 = 16 over 25cosθ= −4 over 5 (negative in Quadrant II)tanθ= 3/5 over −4/5 = −3 over 4cosθ = −4 over 5 and tanθ = −3 over 4. The numbers 3, 4, 5 are a Pythagorean triple, which is why the answer is a clean fraction.
Example 2: cosine given, Quadrant IV
cosθ = 5 over 13 and θ is in Quadrant IV. Find sinθ and tanθ.
sin2θ= 1 − 25 over 169 = 144 over 169sinθ= −12 over 13 (negative in Quadrant IV)tanθ= −12/13 over 5/13 = −12 over 5sinθ = −12 over 13 and tanθ = −12 over 5, from the 5-12-13 triangle.
Example 3: tangent given, Quadrant III
tanθ = 2 and θ is in Quadrant III. Find cosθ and sinθ.
Start with the second identity, then take reciprocals:
sec2θ= 1 + tan2θ = 1 + 4 = 5secθ= −√5 (cosine, and so secant, is negative in Quadrant III)cosθ= −1 over √5 = −√5 over 5 ≈ −0.4472sinθ= tanθ · cosθ = −2√5 over 5 ≈ −0.8944cosθ = −√5 over 5 and sinθ = −2√5 over 5. Check: 5 over 25 + 20 over 25 = 1.
Example 4: a decimal sine
sinθ = 0.28 for an acute angle θ. Find cosθ.
cosθ = √1 − 0.282 = √1 − 0.0784 = √0.9216 = 0.96cosθ = 0.96, positive because acute angles are in Quadrant I. These values come from the 7-24-25 triangle, since 7 over 25 = 0.28 and 24 over 25 = 0.96.
Simplifying and proving with the identities
The identities let you swap sin2θ for 1 − cos2θ (and so on) whenever that makes an expression simpler. Three patterns come up again and again.
Example 5: simplify a fraction
Simplify 1 − cos2x over sinx, assuming sinx ≠ 0.
1 − cos2x over sinx = sin2x over sinx = sinxThe expression equals sinx.
Example 6: expand a square
Expand (sinθ + cosθ)2.
(sinθ + cosθ)2= sin2θ + 2sinθcosθ + cos2θ= 1 + 2sinθcosθThe two squared terms collapse to 1, leaving 1 + 2sinθcosθ.
Example 7: prove an identity
Show that cos4θ − sin4θ = cos2θ − sin2θ.
Factor the left side as a difference of squares, then use the identity on the second factor:
cos4θ − sin4θ= (cos2θ − sin2θ)(cos2θ + sin2θ)= (cos2θ − sin2θ) · 1The two sides are equal for every θ.
The identity also appears in other formulas on this site. The proof of the law of cosines uses it to collapse a2sin2C + a2cos2C into a2, and in physics and engineering it is the reason a vector of length r at angle θ has components rcosθ and rsinθ whose squares add back up to r2. That is the distance formula in polar form.
Common mistakes
- Reading sin2θ as sin(θ2). At θ = 30°, sin230° = 0.25, while sin(900°) = 0. The exponent applies to the value of the sine.
- Thinking sinθ + cosθ = 1. Only the squares add to 1. At 45°, sinθ + cosθ = √2 ≈ 1.414.
- Forgetting the ±. √1 − sin2θ is never negative, so it gives the right answer only in quadrants where cosine is positive. Always choose the sign from the quadrant.
- Splitting a square root. √1 − sin2θ is not 1 − sinθ. With sinθ = 0.6, the first is 0.8 and the second is 0.4.
- Dividing by zero. 1 + tan2θ = sec2θ means nothing at 90° or 270°, where cosθ = 0 and tangent is undefined.
If you are still deciding whether a problem needs trigonometry at all, see Pythagorean theorem or trigonometry?
Frequently asked questions
What are the three Pythagorean identities?
They are sin2θ + cos2θ = 1, 1 + tan2θ = sec2θ and 1 + cot2θ = csc2θ. The second and third come from dividing the first by cos2θ and by sin2θ.
Why is it called the Pythagorean identity?
Because it is the Pythagorean theorem applied to a right triangle with hypotenuse 1. The legs of that triangle are cosθ and sinθ (up to sign), so the theorem a2 + b2 = c2 becomes cos2θ + sin2θ = 1.
Is sin²θ + cos²θ = 1 true for every angle?
Yes, for every real angle: acute, obtuse, negative or larger than 360°. On the unit circle every angle gives a point (cosθ, sinθ) at distance 1 from the origin, and squaring removes any negative signs.
How do you find cos θ if you know sin θ?
Use cosθ = ±√1 − sin2θ, then pick the sign from the quadrant: positive in Quadrants I and IV, negative in II and III. For sinθ = 0.6 in Quadrant II, cosθ = −√1 − 0.36 = −0.8.
Does the Pythagorean identity work in radians?
Yes. The identity is about the point on the unit circle, not about the unit used to measure the angle. For example, sin2π over 6 + cos2π over 6 = 1 over 4 + 3 over 4 = 1, the same as for 30°.
Keep exploring
- LearnEquation of a Circle and the Pythagorean TheoremWhy the equation of a circle, x² + y² = r², is the Pythagorean theorem, plus the center-radius form, worked examples and the inside/outside test.
- ProofsTrigonometric Proofs of the Pythagorean TheoremCan trigonometry prove the Pythagorean theorem without circular reasoning? Jason Zimba's 2009 proof and the Jackson–Johnson law-of-sines proof, explained.
- LearnThe Law of CosinesThe law of cosines c² = a² + b² − 2ab·cos C explained: how it extends the Pythagorean theorem, a proof by dropping an altitude, and SAS and SSS examples.
- LearnThe 30-60-90 TriangleThe 30-60-90 triangle ratio 1 : √3 : 2 derived from half an equilateral triangle, rules for each given side, worked examples and exact values for 30° and 60°.
- LearnPythagorean Theorem or Trigonometry: Which Should You Use?When to use the Pythagorean theorem vs trigonometry (SOH CAH TOA): a decision flowchart, side-by-side worked examples and multi-step problems that need both.
- LearnThe 45-45-90 TriangleThe 45-45-90 triangle rules: why the sides are in the ratio 1 : 1 : √2, how to find a leg from the hypotenuse, the diagonal of a square, and worked examples.