Chapter 3: Pythagorean Identity
Overview
The fundamental trigonometric identity $\sin^2\theta + \cos^2\theta = 1$ is the most basic and important relation among trigonometric functions. It is also called the Pythagorean identity.
This page adopts the unit-circle definition of trigonometric functions. Under this definition, the identity follows immediately, and it is clear that it holds for every real number $\theta$.
Unit-Circle Definition and the Identity
Definition of Trigonometric Functions
Figure 1: The unit circle and the definition of trigonometric functions
Definition: Trigonometric Functions (Unit Circle)
On the unit circle $x^2 + y^2 = 1$, let $P = (x, y)$ be the point reached by rotating an angle $\theta$ counterclockwise from the positive $x$-axis. Then $\cos\theta$ and $\sin\theta$ are defined as follows:
$$\cos\theta \triangleq x, \quad \sin\theta \triangleq y$$This definition applies to every real number $\theta$.
The Fundamental Identity
Since the point $P = (\cos\theta, \sin\theta)$ lies on the unit circle $x^2 + y^2 = 1$, the definition immediately gives:
Pythagorean Identity
$$\sin^2\theta + \cos^2\theta = 1$$This holds for every real number $\theta$.
Key point: Under the unit-circle definition, the Pythagorean identity requires no "proof." The unit-circle equation $x^2 + y^2 = 1$ is itself the statement $\cos^2\theta + \sin^2\theta = 1$.
Right-Triangle Interpretation (Acute Angles)
For $0 < \theta < \frac{\pi}{2}$ (acute angles), the unit-circle definition can be interpreted in terms of the side ratios of a right triangle. This viewpoint provides geometric intuition.
Figure 2: Right triangle ABC (∠B = 90°)
Correspondence with the Right Triangle
In a right triangle with hypotenuse $c$, side $a$ adjacent to angle $\theta$, and side $b$ opposite to $\theta$:
\begin{align} \sin\theta &= \frac{\color{#4CAF50}{b}}{\color{#E91E63}{c}} = \frac{\color{#4CAF50}{\text{opposite}}}{\color{#E91E63}{\text{hypotenuse}}} \\ \cos\theta &= \frac{\color{#FF9800}{a}}{\color{#E91E63}{c}} = \frac{\color{#FF9800}{\text{adjacent}}}{\color{#E91E63}{\text{hypotenuse}}} \end{align}This is consistent with the unit-circle definition. On the unit circle the radius (hypotenuse) is $1$, so $\cos\theta$ and $\sin\theta$ are simply the lengths of the adjacent and opposite sides, respectively.
Derivation from the Pythagorean Theorem
Applying the Pythagorean theorem $a^2 + b^2 = c^2$ to the right triangle:
\begin{eqnarray*} \sin^2\theta + \cos^2\theta &=& \left(\frac{b}{c}\right)^2 + \left(\frac{a}{c}\right)^2 \\ &=& \frac{a^2 + b^2}{c^2} = \frac{c^2}{c^2} = 1 \end{eqnarray*}This explains why the identity is called the "Pythagorean identity."
Note: The right-triangle interpretation applies directly to acute angles $0 < \theta < \frac{\pi}{2}$. For general angles the signs of the coordinates in each quadrant must be considered, but because $\sin^2\theta + \cos^2\theta$ involves squares, the result is the same.
Worked Examples
Example 1
Problem
Given $\sin\theta = \dfrac{3}{5}$ with $0 < \theta < \dfrac{\pi}{2}$, find $\cos\theta$ and $\tan\theta$.
Step 1: Find cos θ
Substitute $\sin\theta = \dfrac{3}{5}$ into the Pythagorean identity $\sin^2\theta + \cos^2\theta = 1$:
$$\left(\frac{3}{5}\right)^2 + \cos^2\theta = 1$$ $$\frac{9}{25} + \cos^2\theta = 1$$ $$\cos^2\theta = 1 - \frac{9}{25} = \frac{16}{25}$$Since $0 < \theta < \dfrac{\pi}{2}$, we have $\cos\theta > 0$, so:
$$\cos\theta = \frac{4}{5}$$Step 2: Find tan θ
From $\tan\theta = \dfrac{\sin\theta}{\cos\theta}$:
$$\tan\theta = \frac{3/5}{4/5} = \frac{3}{4}$$Answer
$$\cos\theta = \frac{4}{5}, \quad \tan\theta = \frac{3}{4}$$Example 2
Problem
Given $\cos\theta = -\dfrac{5}{13}$ with $\pi < \theta < \dfrac{3\pi}{2}$ (third quadrant), find $\sin\theta$ and $\tan\theta$.
Step 1: Find sin θ
From the Pythagorean identity:
$$\sin^2\theta = 1 - \cos^2\theta = 1 - \left(-\frac{5}{13}\right)^2 = 1 - \frac{25}{169} = \frac{144}{169}$$In the third quadrant $\sin\theta < 0$, so:
$$\sin\theta = -\frac{12}{13}$$Step 2: Find tan θ
$$\tan\theta = \frac{\sin\theta}{\cos\theta} = \frac{-12/13}{-5/13} = \frac{12}{5}$$Answer
$$\sin\theta = -\frac{12}{13}, \quad \tan\theta = \frac{12}{5}$$Example 3
Problem
Express $\sin^4\theta + \cos^4\theta$ in terms of $\sin^2\theta\cos^2\theta$.
Solution
Square both sides of $\sin^2\theta + \cos^2\theta = 1$:
$$(\sin^2\theta + \cos^2\theta)^2 = 1^2$$ $$\sin^4\theta + 2\sin^2\theta\cos^2\theta + \cos^4\theta = 1$$Therefore:
$$\sin^4\theta + \cos^4\theta = 1 - 2\sin^2\theta\cos^2\theta$$Answer
$$\sin^4\theta + \cos^4\theta = 1 - 2\sin^2\theta\cos^2\theta$$Practice Problems
Problem 1
Given $\sin\theta = \dfrac{5}{13}$ with $\dfrac{\pi}{2} < \theta < \pi$ (second quadrant), find $\cos\theta$ and $\tan\theta$.
Hint
Use $\sin^2\theta + \cos^2\theta = 1$ to find $\cos^2\theta$. Pay attention to the sign of $\cos\theta$ in the second quadrant.
Solution
$\cos^2\theta = 1 - \sin^2\theta = 1 - \dfrac{25}{169} = \dfrac{144}{169}$
In the second quadrant $\cos\theta < 0$, so $\cos\theta = -\dfrac{12}{13}$
$\tan\theta = \dfrac{\sin\theta}{\cos\theta} = \dfrac{5/13}{-12/13} = -\dfrac{5}{12}$
Problem 2
Given $\tan\theta = 2$ with $0 < \theta < \dfrac{\pi}{2}$, find $\sin\theta$ and $\cos\theta$.
Hint
Use the identity $1 + \tan^2\theta = \dfrac{1}{\cos^2\theta}$ to find $\cos^2\theta$, then compute $\sin\theta = \tan\theta \cdot \cos\theta$.
Solution
Substituting $\tan\theta = 2$ into $1 + \tan^2\theta = \dfrac{1}{\cos^2\theta}$:
$1 + 4 = \dfrac{1}{\cos^2\theta}$, so $\cos^2\theta = \dfrac{1}{5}$
In the first quadrant, $\cos\theta = \dfrac{1}{\sqrt{5}} = \dfrac{\sqrt{5}}{5}$
$\sin\theta = \tan\theta \cdot \cos\theta = 2 \cdot \dfrac{\sqrt{5}}{5} = \dfrac{2\sqrt{5}}{5}$
Problem 3
Express $\sin^6\theta + \cos^6\theta$ in terms of $\sin^2\theta\cos^2\theta$.
Hint
Apply the factorization $a^3 + b^3 = (a + b)(a^2 - ab + b^2)$ with $a = \sin^2\theta$ and $b = \cos^2\theta$. The result of Example 3 may also be useful.
Solution
Let $a = \sin^2\theta$, $b = \cos^2\theta$, so $a + b = 1$:
$a^3 + b^3 = (a + b)(a^2 - ab + b^2) = 1 \cdot (a^2 + b^2 - ab)$
Substituting $a^2 + b^2 = (a+b)^2 - 2ab = 1 - 2ab$:
$= (1 - 2ab) - ab = 1 - 3ab$
$= 1 - 3\sin^2\theta\cos^2\theta$
Problem 4
Prove that $\dfrac{\sin\theta}{1 + \cos\theta} + \dfrac{1 + \cos\theta}{\sin\theta} = \dfrac{2}{\sin\theta}$ (where $\sin\theta \neq 0$ and $\cos\theta \neq -1$).
Hint
Combine the left-hand side into a single fraction by finding a common denominator. After expanding the numerator, apply the Pythagorean identity $\sin^2\theta + \cos^2\theta = 1$.
Solution
Combine the left-hand side over a common denominator:
$\dfrac{\sin\theta}{1 + \cos\theta} + \dfrac{1 + \cos\theta}{\sin\theta} = \dfrac{\sin^2\theta + (1 + \cos\theta)^2}{\sin\theta(1 + \cos\theta)}$
Expand the numerator:
$\sin^2\theta + 1 + 2\cos\theta + \cos^2\theta = (\sin^2\theta + \cos^2\theta) + 1 + 2\cos\theta = 2 + 2\cos\theta = 2(1 + \cos\theta)$
Therefore:
$\dfrac{2(1 + \cos\theta)}{\sin\theta(1 + \cos\theta)} = \dfrac{2}{\sin\theta}$ (matches the right-hand side)