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Algebra 2 Trigonometric functions (introduction)

Pythagorean identity (sin² + cos² = 1)

20 practice questions 0 video lessons Theory + worked examples

The Pythagorean Identity

Common Core Algebra 2 • Standard F-TF.8 • Trigonometric Functions

The Pythagorean Identity is a topic in Trigonometric Functions in the Common Core State Standards. It is aligned to Standard F-TF.8, which requires students to prove the Pythagorean identity and use it to find sine, cosine, or tangent given one value and the quadrant.

The Pythagorean identity \(\sin^2\theta+\cos^2\theta=1\) comes from the unit circle and finds one ratio from the other by quadrant.

Common Core Algebra 2 › Trigonometric Functions › The Pythagorean Identity  —  Standard F-TF.8

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Theory

The Pythagorean identity follows from the right triangle on the unit circle:

\[\sin^2\theta+\cos^2\theta=1.\]

It rearranges to \(1-\sin^2\theta=\cos^2\theta\) and \(1-\cos^2\theta=\sin^2\theta\).

Use the quadrant to choose the sign of the square root.
The Pythagorean identity The right triangle on the unit circle gives sin squared plus cos squared equals 1. cos θ sin θ 1
The unit-circle triangle gives \(\sin^2+\cos^2=1\).
Pythagorean identity Pythagorean identity Pythagorean identity sin²θ + cos²θ = 1 1 - sin²θ = cos²θ 1 - cos²θ = sin²θ
The Pythagorean identity.

The identity:

\[\sin^2\theta+\cos^2\theta=1\]
sine squared theta plus cosine squared theta equals one
Solve for the unknown ratio, then pick the sign by quadrant.

How to use the identity

  1. Substitute the known ratio into \(\sin^2\theta+\cos^2\theta=1\).
  2. Solve for the square of the unknown.
  3. Take the square root.
  4. Choose the sign using the quadrant.
Example 1 — Find cosine
If \(\sin\theta=\dfrac35\) and \(\theta\) is acute, find \(\cos\theta\).
Solution

Use \(\sin^2\theta+\cos^2\theta=1\).

\(\cos^2\theta\)\(=\)\(1-\dfrac{9}{25}=\dfrac{16}{25}\)
\(\cos\theta\)\(=\)\(\dfrac45\)
cosine is four fifths
Example 2 — Second quadrant
If \(\cos\theta=-\dfrac12\) with \(\theta\) in quadrant II, find \(\sin\theta\).
Solution

Solve for \(\sin\theta\); it is positive in quadrant II.

\(\sin^2\theta\)\(=\)\(1-\dfrac14=\dfrac34\)
\(\sin\theta\)\(=\)\(\dfrac{\sqrt3}{2}\)
sine is root 3 over 2
Example 3 — Simplify
Simplify \(1-\sin^2\theta\).
Solution

Rearrange the identity.

\(1-\sin^2\theta\)\(=\)\(\cos^2\theta\)
it equals cosine squared theta
Example 4 — Verify at 30°
Check the identity at \(\theta=30^\circ\).
Solution

\(\sin 30^\circ=\dfrac12,\ \cos 30^\circ=\dfrac{\sqrt3}{2}\).

\(\left(\dfrac12\right)^2+\left(\dfrac{\sqrt3}{2}\right)^2\)\(=\)\(\dfrac14+\dfrac34=1\)
the identity holds, giving 1

Common pitfalls

\(\sin^2\theta\) means \((\sin\theta)^2\), not \(\sin(\theta^2)\).
Choose the sign by quadrant after taking the root.
The identity is always \(=1\), for every angle.

Frequently asked questions

What is the Pythagorean identity?

\(\sin^2\theta+\cos^2\theta=1\).

Where does it come from?

The right triangle on the unit circle, with hypotenuse 1.

How do you find cosine from sine?

\(\cos\theta=\pm\sqrt{1-\sin^2\theta}\), sign by quadrant.

Does \(\sin^2\theta\) mean \(\sin(\theta^2)\)?

No — it means \((\sin\theta)^2\).