3. Trigonometry

e. Inverse Trig Functions

2. Inverse Sine & Cosine

Arc Sine

The sine function is not one-to-one, since for example \[ \sin(\pi-\theta)=\sin(\theta) \] We pick the branch with \(-\,\dfrac{\pi}{2} \le \theta \le \dfrac{\pi}{2}\).

The inverse function of \(\sin\) is \(\arcsin\) (read “arc sine”) or \(\sin^{-1}\) (read “inverse sine”) which satisfies \[ \arcsin(z)=\theta \qquad \text{where} \qquad z=\sin(\theta) \] provided \(-1 \le z \le 1\) and \(-\,\dfrac{\pi}{2} \le \theta \le \dfrac{\pi}{2}\).

Notice that \(-\,\dfrac{\pi}{2} \le \theta \le \dfrac{\pi}{2}\) is quadrants IV and I only.

Compute each of the following.

\(\arcsin\dfrac{1}{2}\)

\(\arcsin\dfrac{1}{2}=\dfrac{\pi}{6}\)
\(\arcsin\dfrac{-1}{\sqrt{2}}\)

\(\arcsin\dfrac{-1}{\sqrt{2}}=-\,\dfrac{\pi}{4}\)
\(\arcsin\dfrac{\sqrt{3}}{2}\)

\(\arcsin\dfrac{\sqrt{3}}{2}=\dfrac{\pi}{3}\)
\(\arcsin(-1)\)

\(\arcsin(-1)=-\,\dfrac{\pi}{2}\)

Arc Cosine

The cosine function is not one-to-one, since for example \[ \cos(-\theta)=\cos(\theta) \] We pick the branch with \(0 \le \theta \le \pi\).

The inverse function of \(\cos\) is \(\arccos\) (read “arc cosine”) or \(\cos^{-1}\) (read “inverse cosine”) which satisfies \[ \arccos(z)=\theta \qquad \text{where} \qquad z=\cos(\theta) \] provided \(-1 \le z \le 1\) and \(0 \le \theta \le \pi\).

Notice that \(0 \le \theta \le \pi\) is quadrants I and II only.