Inverse Trig Expressions

The last property of inverse functions listed on the previous page suggests that if we compose, or combine, a trig function with its inverse trig function, the functions should cancel out. This leads to the following cancellation properties.

Inverse Cancellation Properties: The following properties only apply if the corresponding domain inequalities are satisfied.

\[\begin{align*} \sin^{-1}\left(\sin(x)\right) &= x \text{ if } -\frac{\pi}{2} \le x \le \frac{\pi}{2} \\ \cos^{-1}\left(\cos(x)\right) &= x \text{ if } 0 \le x \le \pi \\ \tan^{-1}\left(\tan(x)\right) &= x \text{ if } -\frac{\pi}{2} \lt x \lt \frac{\pi}{2} \end{align*}\]

\[\begin{align*} \sin\left(\sin^{-1}(x)\right) &= x \text{ if } -1 \le x \le 1 \\ \cos\left(\cos^{-1}(x)\right) &= x \text{ if } -1 \le x \le 1 \\ \tan\left(\tan^{-1}(x)\right) &= x \text{ if } -\infty \lt x \lt \infty \end{align*}\]

Let's also look at some examples that mix the functions being comgined together. We will not be able to apply the cancellation properties with these examples.

Self-Check #7: Evaluate the expression \(\cos^{-1}\left(\cos\left(\frac{5\pi}{4}\right)\right)\). (Select the most appropriate response.)

(A) \(-\frac{\pi}{4}\) (B) \(\frac{\pi}{4}\) (C) \(\frac{3\pi}{4}\) (D) \(\frac{5\pi}{4}\)

You must select an answer above.

(Answer: C) -- If we can apply the cancellation formula, then \(\cos^{-1}\left(\cos\left(\frac{5\pi}{4}\right)\right) = \frac{5\pi}{4}\). Unfortunately, \(\frac{5\pi}{4}\) is not within the interval \(0 \le x \le \pi\). So the cancellation property cannot be used. Instead, we will need to work through the expression from the inside out. From the unit circle, we know that \(\cos\left(\frac{5\pi}{4}\right) = -\frac{\sqrt{2}}{2}\) in the 3^rd quadrant. That simplifies the given expression to \(\cos^{-1}\left(\cos\left(\frac{5\pi}{4}\right)\right) = \cos^{-1}\left(-\frac{\sqrt{2}}{2}\right)\). What angle \(\theta\) between \(0\) and \(\pi\) satisfies the equation \(\cos(\theta) = -\frac{\sqrt{2}}{2}\)? Using the unit circle again, we know that \(\cos\left(\frac{3\pi}{4}\right) = -\frac{\sqrt{2}}{2}\). Thus we can conclude that \(\cos^{-1}\left(\cos\left(\frac{5\pi}{4}\right)\right) = \frac{3\pi}{4}\).

Self-Check #8: Evaluate the expression \(\tan\left(\sin^{-1}\left(-\frac{\sqrt{3}}{2}\right)\right)\). (Select the most appropriate response.)

(A) \(-\sqrt{3}\) (B) \(-\frac{1}{\sqrt{3}}\) (C) \(\frac{1}{\sqrt{3}}\) (D) \(\sqrt{3}\)

You must select an answer above.

(Answer: A) -- We cannot use a cancellation property because the given trig functions are not inverses of each other. So, we will need to work through this from the inside out. Starting with the inside expression \(\sin^{-1}\left(-\frac{\sqrt{3}}{2}\right)\), we need to find an angle between \(-\frac{\pi}{2}\) and \(\frac{\pi}{2}\) that makes \(\sin(\theta) = -\frac{\sqrt{3}}{2}\). Since sign is negative in the 4^th quadrant, we get \(\theta = -\frac{\pi}{3}\). This simplifies the given expression to \(\tan\left(\sin^{-1}\left(-\frac{\sqrt{3}}{2}\right)\right) = \tan\left(-\frac{\pi}{3}\right)\). From the unit circle, we know that \(\tan\left(-\frac{\pi}{3}\right) = -\sqrt{3}\). So we can conclude that \(\tan\left(\sin^{-1}\left(-\frac{\sqrt{3}}{2}\right)\right) = -\sqrt{3}\).

Inverse Trig Functions

Inverse Trig Expressions