AP Calculus AB and BC glossary

Pythagorean Identity

Also called: Pythagorean identities, sin^2 + cos^2 = 1

The Pythagorean identity says that sine squared plus cosine squared equals one, with two more forms found by dividing that equation through by cosine squared or by sine squared.

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

The identity is the unit circle written as algebra. A point on that circle has coordinates $(\cos\theta, \sin\theta)$ and sits one unit from the origin, so the squares of the coordinates add to $1$ . Divide the whole equation by $\cos^2\theta$ , then by $\sin^2\theta$ , and the two forms you meet inside integrals fall out.

1 + \tan^2\theta = \sec^2\theta \qquad 1 + \cot^2\theta = \csc^2\theta

Rewriting is the whole job. $\int \tan^2 x\,dx$ has no rule of its own until $\tan^2 x$ becomes $\sec^2 x - 1$ , and then it integrates on sight to $\tan x - x + C$ . The identity also explains how two people can hand in different antiderivatives and both be right: substituting $u = \sin x$ into $\int \sin x\cos x\,dx$ gives $\tfrac{1}{2}\sin^2 x + C$ , while $u = \cos x$ gives $-\tfrac{1}{2}\cos^2 x + C$ , and those two expressions differ by exactly $\tfrac{1}{2}$ .

The mistake

It is $\tan^2\theta = \sec^2\theta - 1$ , never $\sec^2\theta + 1$ . What makes the sign slip dangerous is that nothing stalls when you make it: $\int (\sec^2 x + 1)\,dx = \tan x + x + C$ looks every bit as finished as the correct $\tan x - x + C$ , so a wrong answer walks straight onto the page. Rebuild the form from $\sin^2\theta + \cos^2\theta = 1$ when you are unsure, and differentiate your answer back. Inside a trig substitution the same slip does stall you, since $\sqrt{\sec^2\theta - 1}$ simplifies to $|\tan\theta|$ while $\sqrt{\sec^2\theta + 1}$ simplifies to nothing.

Appears in: Unit 6: Integration and Accumulation