AP Calculus AB and BC glossary

Inverse trigonometric function

Also called: Inverse trig function

An inverse trigonometric function undoes a trig function on a domain restricted to make it one to one. Arcsine returns angles from negative pi over two to pi over two, arccosine returns angles from zero to pi, and arctangent returns angles strictly between negative pi over two and pi over two.

sin⁡y=x  ⇒  cos⁡y⋅dydx=1  ⇒  dydx=11−x2\sin y = x \;\Rightarrow\; \cos y \cdot \frac{dy}{dx} = 1 \;\Rightarrow\; \frac{dy}{dx} = \frac{1}{\sqrt{1 - x^{2}}}

Sine takes the value 12\frac{1}{2} at infinitely many angles, so it has no inverse until the domain is cut down. The agreed cut keeps −π2≤θ≤π2-\frac{\pi}{2} \le \theta \le \frac{\pi}{2} for sine and 0≤θ≤π0 \le \theta \le \pi for cosine, which is why arcsin⁡(sin⁡5π6)\arcsin\left(\sin\frac{5\pi}{6}\right) is π6\frac{\pi}{6} and not 5π6\frac{5\pi}{6}.

Writing y=arcsin⁡xy = \arcsin x as sin⁡y=x\sin y = x and differentiating implicitly gives dydx=1cos⁡y\frac{dy}{dx} = \frac{1}{\cos y}, and the Pythagorean identity turns cos⁡y\cos y into 1−x2\sqrt{1 - x^{2}}. The positive root is correct only because the restricted range keeps cos⁡y≥0\cos y \ge 0, so the domain restriction is what fixes the sign. The algebraic result is why these functions show up as standard antiderivatives, as in ∫11+x2 dx=arctan⁡x+C\int \frac{1}{1 + x^{2}}\,dx = \arctan x + C.

The mistake

Reading sin⁡−1x\sin^{-1}x as 1sin⁡x\frac{1}{\sin x}. The −1-1 marks an inverse function, while the 2 in sin⁡2x\sin^{2}x marks a power, so one notation means two different things. The reciprocal of sin⁡x\sin x is csc⁡x\csc x, a completely different function with a completely different derivative.

Appears in: Unit 3: Chain Rule, Implicit, and Inverses, Unit 6: Integration and Accumulation