Pythagorean Identity
Description
The Pythagorean Identity is the single most important equation in trigonometry. It bridges the gap between geometry (the Pythagorean Theorem) and trigonometry (sine and cosine).
It states that for *any* angle $\theta$, the square of the sine of that angle plus the square of the cosine of that angle always equals exactly 1.
This identity is derived directly from the Unit Circle, where any point $(x, y)$ on the circle satisfies the equation $x^2 + y^2 = 1$. Since $x = \cos(\theta)$ and $y = \sin(\theta)$, it follows directly that $\cos^2(\theta) + \sin^2(\theta) = 1$.
History & Origins
While the theorem for triangles is named after Pythagoras (c. 570 BC), the trigonometric form was developed much later as trigonometry evolved from chord functions to sine/cosine. Claudius Ptolemy (c. 100 AD): In his Almagest, he used chord tables that implicitly relied on this relationship (Chord^2 + Supplement^2 = Diameter^2). Indian Mathematicians (c. 500 AD): Aryabhata and later mathematicians explicitly used the relationship $\sin^2 + \cos^2 = 1$ (using their terms jya and koti-jya) for astronomical calculations.
Proof using the Unit Circle
We derive the identity directly from the definition of the unit circle.
Consider a Unit Circle centered at the origin $(0,0)$ with radius $r=1$.
The equation of this circle is $x^2 + y^2 = 1$.
Draw an angle $\theta$ in standard position. The terminal side intersects the circle at point $P(x,y)$.
By the definition of sine and cosine on the unit circle: $x = \cos(\theta)$ and $y = \sin(\theta)$.
Substitute these into the circle equation: $(\cos\theta)^2 + (\sin\theta)^2 = 1$.
This simplifies to the standard notation: $\cos^2\theta + \sin^2\theta = 1$.
Variables
| Symbol | Meaning |
|---|---|
θ | Angle (any real number) |
sin²θ | Square of the sine of the angle |
cos²θ | Square of the cosine of the angle |
Examples
Basic Calculation
Problem: Verify the identity for θ = 30°
Solution:
Finding Missing Value
Problem: If $\sin(\theta) = \frac{3}{5}$ and $\theta$ is in the first quadrant, find $\cos(\theta)$.
Solution: 4/5
- Identity: $\sin^2(\theta) + \cos^2(\theta) = 1$.
- Substitute: $(\frac{3}{5})^2 + \cos^2(\theta) = 1$.
- Square: $\frac{9}{25} + \cos^2(\theta) = 1$.
- Subtract: $\cos^2(\theta) = 1 - \frac{9}{25} = \frac{16}{25}$.
- Square Root: $\cos(\theta) = \pm \sqrt{\frac{16}{25}} = \pm \frac{4}{5}$.
- Since it is in Quadrant I, cosine is positive: $\frac{4}{5}$.
Simplifying Expressions
Problem: Simplify the expression $5 - 5\sin^2(x)$.
Solution: $5\cos^2(x)$
- Factor out 5: $5(1 - \sin^2(x))$.
- Rearrange identity: Since $\sin^2 + \cos^2 = 1$, then $1 - \sin^2 = \cos^2$.
- Substitute: $5(\cos^2(x))$.
Common Mistakes
Thinking sin(x) + cos(x) = 1
This is a very common error. The identity applies to the **squares**, not the linear functions. For example, at $45^\circ$, $\frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} = \sqrt{2} \approx 1.414$, not 1.
Inverse notation confusion
$\sin^2(x)$ means $(\sin(x))^2$. It is NOT the same as $\sin(x^2)$ (squaring the angle) or $\sin^{-1}(x)$ (inverse sine).
Real-World Applications
Electrical Engineering
In AC circuits, the relationship between Active Power (P), Reactive Power (Q), and Apparent Power (S) forms a "Power Triangle" governed by this identity ($S^2 = P^2 + Q^2$). It is crucial for calculating power factors.
Physics: Simple Harmonic Motion
When a pendulum swings or a spring oscillates, the kinetic energy (related to velocity, cosine) and potential energy (related to position, sine) constantly transform into each other. The total energy remains constant, mirroring $\sin^2 + \cos^2 = 1$.
Frequently Asked Questions
Why do we write sin²θ instead of (sin θ)²?
It is a mathematical convention to avoid ambiguity. If we wrote $\sin \theta^2$, it might look like we are squaring the angle $\theta$. Writing the 2 next to the sin makes it clear we are squaring the function result.
Does this work for angles greater than 90°?
Yes, it works for ALL real numbers, including negative angles and angles larger than 360°, because the unit circle definition ($x^2+y^2=1$) is always true regardless of the quadrant.