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Proving the
Pythagorean Identity

Discover why \(\sin^2\theta + \cos^2\theta = 1\) is always true, and learn to use it to find trig values in any quadrant.

Precalculus

The Unit Circle Foundation

The unit circle is a circle centered at the origin with radius \(r = 1\). Any point on this circle can be written as \((\cos\theta,\, \sin\theta)\), where \(\theta\) is the angle measured from the positive \(x\)-axis.

\((\cos\theta,\,\sin\theta)\)
\(\cos\theta\)
\(\sin\theta\)
\(\theta\)
 O x y
\(r=1\)

From Pythagoras to Trigonometry

Draw a right triangle inside the unit circle. The legs have lengths \(\cos\theta\) and \(\sin\theta\), and the hypotenuse is the radius \(r = 1\). By the Pythagorean theorem:

Since \(r = 1\) on the unit circle, this simplifies directly to the identity:

This holds for every angle \(\theta\): any quadrant, positive or negative, radians or degrees.

A Visual Proof

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

The Pythagorean Identity

This identity can be rearranged into two other useful forms:

Solve for \(\sin^2\theta\)
$$\sin^2\theta = 1 - \cos^2\theta$$
Solve for \(\cos^2\theta\)
$$\cos^2\theta = 1 - \sin^2\theta$$

Two Derived Identities

Dividing \(\sin^2\theta + \cos^2\theta = 1\) by \(\cos^2\theta\) or \(\sin^2\theta\) produces two more Pythagorean identities:

Divide by \(\cos^2\theta\)
$$\tan^2\theta + 1 = \sec^2\theta$$
Divide by \(\sin^2\theta\)
$$1 + \cot^2\theta = \csc^2\theta$$

Identity vs Equation

Identity Equation
Definition True for all valid inputs True for specific values only
Example \(\sin^2\theta + \cos^2\theta = 1\) \(\sin\theta = \tfrac{1}{2}\)
How you use it Substitute freely to simplify Solve for unknown \(\theta\)
Must prove? Yes: show both sides always equal No: just find solutions

Sign Rules by Quadrant

Q I
\(\sin\theta>0, \cos\theta>0, \tan\theta>0\)
 "All" Q II
\(\sin\theta>0, \cos\theta<0, \tan\theta<0\)
 "Students" Q III
\(\sin\theta<0, \cos\theta<0, \tan\theta>0\)
 "Take" Q IV
\(\sin\theta<0, \cos\theta>0, \tan\theta<0\)
 "Calculus"

How to Apply the Identity

1
Identify what you know. Are you given \(\sin\theta\), \(\cos\theta\), or \(\tan\theta\)?
2
Substitute into the identity. Plug the known value into \(\sin^2\theta + \cos^2\theta = 1\).
3
Solve algebraically. Isolate the unknown squared term, then take the square root.
4
Choose the correct sign. Use the given quadrant to decide if the answer is positive or negative.
5
Find other ratios if needed. Use \(\tan\theta = \frac{\sin\theta}{\cos\theta}\) to find remaining values.

Example 1: Find \(\cos\theta\)

Given: \(\sin\theta = \dfrac{3}{5}\) and \(\theta\) is in Quadrant I. Find \(\cos\theta\).

1
Start with the identity: \(\sin^2\theta + \cos^2\theta = 1\)
2
Substitute: \(\left(\dfrac{3}{5}\right)^2 + \cos^2\theta = 1\)
3
Simplify: \(\dfrac{9}{25} + \cos^2\theta = 1\), so \(\cos^2\theta = \dfrac{16}{25}\)
4
Square root: \(\cos\theta = \pm\dfrac{4}{5}\)
5
Choose sign: Q I means cosine is positive, so \(\cos\theta = \dfrac{4}{5}\)

Example 2: Find \(\sin\theta\) and \(\tan\theta\)

Given: \(\cos\theta = -\dfrac{5}{13}\) and \(\theta\) is in Quadrant III. Find \(\sin\theta\) and \(\tan\theta\).

1
Substitute: \(\sin^2\theta + \left(-\dfrac{5}{13}\right)^2 = 1\), so \(\sin^2\theta + \dfrac{25}{169} = 1\)
2
Solve: \(\sin^2\theta = \dfrac{144}{169}\), so \(\sin\theta = \pm\dfrac{12}{13}\)
3
Choose sign: In Q III, sine is negative: \(\sin\theta = -\dfrac{12}{13}\)
4
Find tangent: \(\tan\theta = \dfrac{\sin\theta}{\cos\theta} = \dfrac{-12/13}{-5/13} = \dfrac{12}{5}\)

Example 3: Starting from \(\tan\theta\)

Given: \(\tan\theta = -\dfrac{7}{24}\) and \(\theta\) is in Quadrant II. Find \(\sin\theta\) and \(\cos\theta\).

1
Use derived identity: \(\tan^2\theta + 1 = \sec^2\theta\)
2
Substitute: \(\left(-\dfrac{7}{24}\right)^2 + 1 = \sec^2\theta\), so \(\dfrac{49}{576} + 1 = \dfrac{625}{576}\)
3
Solve: \(\sec\theta = \pm\dfrac{25}{24}\). Q II: cosine is negative, so \(\sec\theta = -\dfrac{25}{24}\), thus \(\cos\theta = -\dfrac{24}{25}\)
4
Find sine: \(\sin\theta = \tan\theta \cdot \cos\theta = \left(-\dfrac{7}{24}\right)\left(-\dfrac{24}{25}\right) = \dfrac{7}{25}\)

Simplifying with the Identity

The Pythagorean identity is a powerful tool for simplifying trigonometric expressions. Whenever you see \(\sin^2\theta + \cos^2\theta\), replace it with 1.

Example A
$$\frac{\sin^2\theta + \cos^2\theta}{\cos\theta} = \frac{1}{\cos\theta} = \sec\theta$$
Example B
$$1 - \sin^2\theta = \cos^2\theta$$

Common Mistakes

✕ Forgetting the square root
After finding \(\cos^2\theta = \frac{16}{25}\), students write \(\cos\theta = \frac{16}{25}\) instead of \(\cos\theta = \pm\frac{4}{5}\). Always take the square root.
✕ Wrong sign from quadrant
Solving correctly but choosing the positive root in a quadrant where the value is negative. Always check the quadrant before picking \(+\) or \(-\).
✕ Writing \(\sin^2\theta\) as \(\sin(\theta^2)\)
The notation \(\sin^2\theta\) means \((\sin\theta)^2\), not the sine of \(\theta\) squared. This is a critical distinction.

Summary of Pythagorean Identities

Identity Derived By Useful When
\(\sin^2\theta + \cos^2\theta = 1\) Pythagorean theorem on unit circle Finding \(\sin\theta\) or \(\cos\theta\)
\(\tan^2\theta + 1 = \sec^2\theta\) Divide by \(\cos^2\theta\) Given \(\tan\theta\), find \(\sec\theta\)
\(1 + \cot^2\theta = \csc^2\theta\) Divide by \(\sin^2\theta\) Given \(\cot\theta\), find \(\csc\theta\)

Question 1 of 5

If \(\sin\theta = \dfrac{5}{13}\) and \(\theta\) is in Quadrant II, what is \(\cos\theta\)?

Question 2 of 5

Which of the following is equivalent to \(1 - \cos^2\theta\)?

Question 3 of 5

If \(\cos\theta = \dfrac{3}{5}\) and \(\theta\) is in Quadrant IV, what is \(\tan\theta\)?

Question 4 of 5

Simplify: \(\dfrac{\sin^2\theta}{1 - \cos\theta}\), given that \(\cos\theta \neq 1\).

Question 5 of 5

If \(\tan\theta = 2\) and \(\theta\) is in Quadrant III, what is \(\sin\theta\)?