Algebra 2
Trigonometric functions (introduction)
Pythagorean identity (sin² + cos² = 1)
20 practice questions
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Theory + worked examples
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 unit-circle triangle gives \(\sin^2+\cos^2=1\).
The Pythagorean identity.
The identity:
\[\sin^2\theta+\cos^2\theta=1\]
Solve for the unknown ratio, then pick the sign by quadrant.
How to use the identity
- Substitute the known ratio into \(\sin^2\theta+\cos^2\theta=1\).
- Solve for the square of the unknown.
- Take the square root.
- Choose the sign using the quadrant.
Example 1 — Find cosine
If \(\sin\theta=\dfrac35\) and \(\theta\) is acute, find \(\cos\theta\).
Solution
Use \(\sin^2\theta+\cos^2\theta=1\).
| \(\cos^2\theta\) | \(=\) | \(1-\dfrac{9}{25}=\dfrac{16}{25}\) |
| \(\cos\theta\) | \(=\) | \(\dfrac45\) |
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.
| \(\sin^2\theta\) | \(=\) | \(1-\dfrac14=\dfrac34\) |
| \(\sin\theta\) | \(=\) | \(\dfrac{\sqrt3}{2}\) |
Example 3 — Simplify
Simplify \(1-\sin^2\theta\).
Solution
Rearrange the identity.
| \(1-\sin^2\theta\) | \(=\) | \(\cos^2\theta\) |
Example 4 — Verify at 30°
Check the identity at \(\theta=30^\circ\).
Solution
\(\sin 30^\circ=\dfrac12,\ \cos 30^\circ=\dfrac{\sqrt3}{2}\).
| \(\left(\dfrac12\right)^2+\left(\dfrac{\sqrt3}{2}\right)^2\) | \(=\) | \(\dfrac14+\dfrac34=1\) |
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 \((\sin\theta)^2\).
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