The Pythagorean Identity
Aligned to F-TF.C.8 — Common Core State Standards for Mathematics.
What this lesson teaches
On the unit circle, a point (cos θ, sin θ) lies at distance 1 from the center. The Pythagorean theorem on that radius gives sin^2(θ) + cos^2(θ) = 1, which holds for every angle and lets you find one ratio from another.
Worked examples
Example 1
If sin θ = 3/5 and θ is in the second quadrant, find cos θ. Use cos^2 θ = 1 − sin^2 θ = 1 − 9/25 = 16/25, so cos θ = ±4/5. In the second quadrant cosine is negative, so cos θ = −4/5.
Example 2
If tan θ = 3/4 and θ is in the third quadrant, find sin θ. Tangent is opposite/adjacent, so build a triangle with legs 3 and 4 and hypotenuse √(9 + 16) = 5. That gives |sin θ| = 3/5. In the third quadrant sine is negative, so sin θ = −3/5, and cos θ = −4/5. Check: (−3/5)² + (−4/5)² = 9/25 + 16/25 = 1 ✓.
Example 3
Common mistake: taking only the positive square root. Solving cos²θ = 16/25 gives cos θ = ±4/5 — BOTH signs are mathematically available, and only the quadrant decides between them. Dropping the ± and writing 4/5 silently assumes the first quadrant. Always write both, then use the quadrant information the problem gave you to pick one.
Practice questions
- If cos θ = 5/13 in the first quadrant, find sin θ.
- Verify sin^2(30°) + cos^2(30°) = 1.
- If sin θ = −1/2 in the third quadrant, find cos θ.
Watch the lesson
Every lesson comes with a video taught in English and Spanish — the same video the QR code in the printed workbook opens.
▶ Watch this lessonEn español
La Identidad Pitagórica
En el círculo unitario, un punto (cos θ, sen θ) está a distancia 1 del centro. El teorema de Pitágoras sobre ese radio da sen^2(θ) + cos^2(θ) = 1, que se cumple para todo ángulo y permite hallar una razón a partir de otra.
Ejemplo 1: Si sen θ = 3/5 y θ está en el segundo cuadrante, halla cos θ. Usa cos^2 θ = 1 − sen^2 θ = 1 − 9/25 = 16/25, así que cos θ = ±4/5. En el segundo cuadrante el coseno es negativo, por lo que cos θ = −4/5.
Ejemplo 2: Si tan θ = 3/4 y θ está en el tercer cuadrante, halla sen θ. La tangente es opuesto/adyacente, así que arma un triángulo con catetos 3 y 4 e hipotenusa √(9 + 16) = 5. Eso da |sen θ| = 3/5. En el tercer cuadrante el seno es negativo, así que sen θ = −3/5 y cos θ = −4/5. Comprueba: (−3/5)² + (−4/5)² = 9/25 + 16/25 = 1 ✓.
Ejemplo 3: Error común: tomar solo la raíz cuadrada positiva. Resolver cos²θ = 16/25 da cos θ = ±4/5 — AMBOS signos están matemáticamente disponibles, y solo el cuadrante decide entre ellos. Quitar el ± y escribir 4/5 supone en silencio el primer cuadrante. Escribe siempre ambos y luego usa la información del cuadrante que el problema te dio para elegir uno.
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