Pythagorean Identity

The Pythagorean Identity states that sine squared plus cosine squared always equals one, helping you simplify trigonometric expressions and solve equations.

sin⁡2θ+cos⁡2θ=1\sin^2\theta + \cos^2\theta = 1

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What each symbol means

What Pythagorean Identity takes
θ\theta
Pythagorean Identity →
SymbolMeaning
θ\thetaThe angle variable, measured in radians or degrees; represents the input to the sine and cosine functions, and confusing units between them will give an incorrect answer.

When to use it

Use the Pythagorean Identity when you need to rewrite or simplify expressions containing both sine and cosine, or when solving trigonometric equations.

Level

Usually taught in: Algebra II · Appears on: SAT, ACT

Worked examples

1. Find cosine when sine is known

Problem

If sin⁡θ=35\sin\theta = \frac{3}{5} and θ\theta is in the first quadrant, find cos⁡θ\cos\theta.
  1. sin⁡2θ+cos⁡2θ=1\sin^2\theta + \cos^2\theta = 1

    Write the Pythagorean Identity.

  2. (35)2+cos⁡2θ=1\left(\frac{3}{5}\right)^2 + \cos^2\theta = 1

    Substitute sin⁡θ=35\sin\theta = \frac{3}{5} into the identity.

  3. 925+cos⁡2θ=1\frac{9}{25} + \cos^2\theta = 1

    Simplify: (35)2=925\left(\frac{3}{5}\right)^2 = \frac{9}{25}.

  4. cos⁡2θ=1625\cos^2\theta = \frac{16}{25}

    Subtract 925\frac{9}{25} from both sides: 1−925=16251 - \frac{9}{25} = \frac{16}{25}.

  5. cos⁡θ=±45\cos\theta = \pm\frac{4}{5}

    Take the square root of both sides.

  6. cos⁡θ=45\cos\theta = \frac{4}{5}

    Since θ\theta is in the first quadrant, both sine and cosine are positive, so choose the positive solution.

Answer: cos⁡θ=45\cos\theta = \frac{4}{5}

The Pythagorean Identity lets us find the missing trig ratio when we know one. We substitute, solve for the unknown, and use the quadrant to pick the correct sign.

2. Solve for squared cosine with a negative sine

Problem

If sin⁡θ=−23\sin\theta = -\frac{2}{3} and cos⁡θ<0\cos\theta < 0, find cos⁡2θ\cos^2\theta.
  1. sin⁡2θ+cos⁡2θ=1\sin^2\theta + \cos^2\theta = 1

    Write the Pythagorean Identity.

  2. (−23)2+cos⁡2θ=1\left(-\frac{2}{3}\right)^2 + \cos^2\theta = 1

    Substitute sin⁡θ=−23\sin\theta = -\frac{2}{3}.

  3. 49+cos⁡2θ=1\frac{4}{9} + \cos^2\theta = 1

    Simplify: (−23)2=49\left(-\frac{2}{3}\right)^2 = \frac{4}{9} (a negative squared is positive).

  4. cos⁡2θ=1−49=59\cos^2\theta = 1 - \frac{4}{9} = \frac{5}{9}

    Subtract 49\frac{4}{9} from both sides to isolate cos⁡2θ\cos^2\theta.

Answer: cos⁡2θ=59\cos^2\theta = \frac{5}{9}

When you only need the squared value, stop without taking the square root. The given constraints (sin⁡θ\sin\theta negative and cos⁡θ\cos\theta negative) place the angle in the third quadrant, which you can use to verify your answer makes sense.

3. Find a shot angle component using trigonometry

Problem

During basketball season, a player's jump shot can be analyzed using the vertical and horizontal components of its trajectory. If the vertical component of the release angle is sin⁡θ=0.8\sin\theta = 0.8 where 0<θ<π20 < \theta < \frac{\pi}{2}, what is the horizontal component cos⁡θ\cos\theta?
  1. sin⁡2θ+cos⁡2θ=1\sin^2\theta + \cos^2\theta = 1

    The Pythagorean Identity models how the vertical and horizontal components relate.

  2. (0.8)2+cos⁡2θ=1(0.8)^2 + \cos^2\theta = 1

    Substitute sin⁡θ=0.8\sin\theta = 0.8 for the vertical component.

  3. 0.64+cos⁡2θ=10.64 + \cos^2\theta = 1

    Calculate: (0.8)2=0.64(0.8)^2 = 0.64.

  4. cos⁡2θ=0.36\cos^2\theta = 0.36

    Subtract: 1−0.64=0.361 - 0.64 = 0.36.

  5. cos⁡θ=±0.6\cos\theta = \pm 0.6

    Take the square root of both sides.

  6. cos⁡θ=0.6\cos\theta = 0.6

    Since 0<θ<π20 < \theta < \frac{\pi}{2} (first quadrant), cosine is positive.

Answer: cos⁡θ=0.6\cos\theta = 0.6

The Pythagorean Identity applies to real-world problems involving angles and components. Knowing the vertical component of the trajectory lets us find the horizontal component, which helps determine the shot's efficiency and arc.

Common mistakes

Where Pythagorean Identity usually goes wrong
Answer came out wrong
Writing sin⁡θ+cos⁡θ=1\sin\theta + \cos\theta = 1 instead of sin⁡2θ+cos⁡2θ=1\sin^2\theta + \cos^2\theta = 1.
Always include both squared terms: sin⁡2θ+cos⁡2θ=1\sin^2\theta + \cos^2\theta = 1.
Writing cos⁡θ=±1−sin⁡2θ\cos\theta = \pm\sqrt{1 - \sin^2\theta} and leaving the answer with both signs instead of choosing one.
Use the quadrant rules—or the given information in the problem—to select the correct sign, then write a single answer like cos⁡θ=0.6\cos\theta = 0.6 or cos⁡θ=−0.6\cos\theta = -0.6.
Rearranging to cos⁡2θ=1−sin⁡2θ\cos^2\theta = 1 - \sin^2\theta and then writing cos⁡θ=1−sin⁡θ\cos\theta = 1 - \sin\theta.
Isolate the squared term first, then take the square root: cos⁡θ=±1−sin⁡2θ\cos\theta = \pm\sqrt{1 - \sin^2\theta}.
The mistakeWhy it is wrongThe fix
Writing sin⁡θ+cos⁡θ=1\sin\theta + \cos\theta = 1 instead of sin⁡2θ+cos⁡2θ=1\sin^2\theta + \cos^2\theta = 1.Dropping the exponents changes the identity into a different (false) equation that will lead to wrong answers.Always include both squared terms: sin⁡2θ+cos⁡2θ=1\sin^2\theta + \cos^2\theta = 1.
Writing cos⁡θ=±1−sin⁡2θ\cos\theta = \pm\sqrt{1 - \sin^2\theta} and leaving the answer with both signs instead of choosing one.The identity gives two algebraic solutions, but the angle's quadrant or the problem's constraints tell you which sign is correct for that specific angle.Use the quadrant rules—or the given information in the problem—to select the correct sign, then write a single answer like cos⁡θ=0.6\cos\theta = 0.6 or cos⁡θ=−0.6\cos\theta = -0.6.
Rearranging to cos⁡2θ=1−sin⁡2θ\cos^2\theta = 1 - \sin^2\theta and then writing cos⁡θ=1−sin⁡θ\cos\theta = 1 - \sin\theta.You cannot remove the squares by subtracting; a−b≠a−b\sqrt{a - b} \ne \sqrt{a} - \sqrt{b}.Isolate the squared term first, then take the square root: cos⁡θ=±1−sin⁡2θ\cos\theta = \pm\sqrt{1 - \sin^2\theta}.

Tips and when to use something else

  • The identity works for all angles, not just acute angles in the first quadrant—always use quadrant rules or given constraints to pick the correct sign when you take the square root.
  • You can rearrange the identity to sin⁡2θ=1−cos⁡2θ\sin^2\theta = 1 - \cos^2\theta or cos⁡2θ=1−sin⁡2θ\cos^2\theta = 1 - \sin^2\theta, whichever isolates the quantity you need.
  • If you only need sin⁡2θ\sin^2\theta or cos⁡2θ\cos^2\theta (not the unsquared value), stop after solving for the squared term and skip the square root step.
  • For problems involving tangent, use the definition tan⁡θ=sin⁡θcos⁡θ\tan\theta = \frac{\sin\theta}{\cos\theta} or Reciprocal Identities instead of forcing this identity.

Related formulas

Sine Cosine TangentFind missing sides or angles in right triangles using sine, cosine, and tangent ratios—essential for SAT and ACT geometry.Unit CircleThe unit circle is a fundamental tool that shows how angles map to sine and cosine coordinates: the point (cos θ, sin θ) always lies on x² + y² = 1.Reciprocal IdentitiesReciprocal identities convert trig functions to their reciprocals, helping you simplify expressions and solve equations with cosecant, secant, and cotangent.Solving Trigonometric EquationsFind all angles that produce a specific sine value by using inverse sine and the periodic nature of trigonometric functions.

Frequently asked questions

Why is it called the Pythagorean Identity?
The name comes from the Pythagorean theorem: a2+b2=c2a^2 + b^2 = c^2. If you place a right triangle in the unit circle (where c=1c = 1), the legs are sin⁡θ\sin\theta and cos⁡θ\cos\theta, so sin⁡2θ+cos⁡2θ=12=1\sin^2\theta + \cos^2\theta = 1^2 = 1. This connection shows why the identity is named after Pythagoras.
Can I use the Pythagorean Identity if my angle is in degrees instead of radians?
Yes, the identity works whether your angle is in degrees or radians. What matters is that you are consistent throughout your work: use the same unit for all angles, and set your calculator to the same mode (degree or radian) when evaluating sine and cosine functions.
What should I do if I get a negative number under the square root?
That means you have made an arithmetic error earlier in your work. The expression 1−sin⁡2θ1 - \sin^2\theta (or 1−cos⁡2θ1 - \cos^2\theta) must always be non-negative because sin⁡2θ\sin^2\theta and cos⁡2θ\cos^2\theta are always between 0 and 1. Retrace your steps to find and fix the mistake.
How do I know whether to use the positive or negative square root?
The sign depends on the angle's quadrant: first quadrant (both positive), second quadrant (sine positive, cosine negative), third quadrant (both negative), and fourth quadrant (cosine positive, sine negative). Check the problem for this information, or use the given constraints to figure out which quadrant you are in, then pick the correct sign.

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Reviewed 2026-09-18