AP Calculus AB and BC glossary

Double-Angle Identity

Also called: Power-reducing identity, Half-angle formula

A double-angle identity writes a trig function of 2u in terms of functions of u. Read backwards, it writes a squared function as a first power of cos 2u, and those power-reducing forms are the ones calculus needs, because no substitution touches the integral of cosine squared until the square is gone.

\cos 2u = \cos^2 u - \sin^2 u \qquad \sin 2u = 2\sin u\cos u

Everything comes from $\cos 2u = \cos^2 u - \sin^2 u$ . Replace one square using the Pythagorean identity and you get $\cos 2u = 2\cos^2 u - 1$ , replace the other and you get $\cos 2u = 1 - 2\sin^2 u$ . Solving each of those for the square is the step that matters going into an integral.

\cos^2 u = \frac{1 + \cos 2u}{2} \qquad \sin^2 u = \frac{1 - \cos 2u}{2}

No substitution touches $\int \cos^2 x\,dx$ as it stands, since the derivative of the inside function is nowhere in the integrand. Integration by parts can grind it out, but power reduction is the route the AP course expects: rewrite the integral as $\int \frac{1 + \cos 2x}{2}\,dx$ and it splits into two terms you already know, giving $\frac{x}{2} + \frac{\sin 2x}{4} + C$ . The identities work in the other direction too. $\sin 2u = 2\sin u\cos u$ collapses a product into one term, and substituting $u = \theta/2$ into the power-reducing form gives the half-angle formula $\cos\frac{\theta}{2} = \pm\sqrt{\frac{1 + \cos\theta}{2}}$ , with the sign set by the quadrant of $\theta/2$ .

The mistake

Losing the inside factor of $2$ once the rewriting is done. The antiderivative of $\cos 2x$ is $\frac{1}{2}\sin 2x$ , so the term above carries a quarter rather than a half. Differentiate whatever you write down; the chain rule hands the missing factor straight back to you.

Appears in: Unit 6: Integration and Accumulation