12.1 Inverse Trigonometric Values

The following inverse trigonometric expressions are all equal to $$\frac{\pi}{4}$$ :

$$\sin^{-1}\left(\frac{\sqrt{2}}{2}\right)$$

$$\cos^{-1}\left(\frac{\sqrt{2}}{2}\right)$$

$$\tan^{-1}\left(1\right)$$

The interval $$\left[-\frac{\pi}{2},\ \frac{\pi}{2}\right]$$ is a continuous interval that covers all sine values.

Restricting the domain of $$y=\sin x$$ to $$\left[-\frac{\pi}{2},\ \frac{\pi}{2}\right]$$ makes the range of the inverse sine function $$y=\sin^{-1}x$$ also $$\left[-\frac{\pi}{2},\ \frac{\pi}{2}\right]$$ . All inverse sine values can only be from $$\left[-\frac{\pi}{2},\ \frac{\pi}{2}\right]$$ .

The range of the inverse cosine function $$y=\cos^{-1}x$$ is the restricted domain of the cosine function, $$\left[0,\ \pi\right]$$ .

This means that when evaluating an inverse cosine function, the resulting angle can only be from the interval $$\left[0,\ \pi\right]$$ .

The restriction for the tangent function is the open interval $$\left(-\frac{\pi}{2},\ \frac{\pi}{2}\right)$$ instead of the closed interval $$\left[-\frac{\pi}{2},\ \frac{\pi}{2}\right]$$ because tangent values are undefined at $$\pm\frac{\pi}{2}$$ .

Inverse tangent values are limited to $$\left(-\frac{\pi}{2},\ \frac{\pi}{2}\right)$$ .

First identify the angle in Quadrant I that would make sine equal to $$\frac{\sqrt{3}}{2}$$ .

This angle is $$\frac{\pi}{3}$$ .

The angle that would give a sine value of $$-\frac{\sqrt{3}}{2}$$ is $$-\frac{\pi}{3}$$ .

We can apply the same technique since the range of inverse sine and the range of inverse tangent cover the same quadrants.

From Quadrant I, we have $$\tan^{-1}\left(\frac{\sqrt{3}}{3}\right)\ =\ \frac{\pi}{6}$$ .

This translates to Quadrant IV as

$$\tan^{-1}\left(-\frac{\sqrt{3}}{3}\right)\ =\ -\frac{\pi}{6}$$ .

The angle is in Quadrant II since the cosine value is negative. Visually, a cosine value of $$-\frac{1}{2}$$ means the angle is closer to the y-axis. This angle is $$\frac{2\pi}{3}$$ .

Note: The technique previously used for inverse sine and inverse tangent can no longer be applied to inverse cosine because its range covers a different pair of quadrants.

For a cosine value of $$-\frac{\sqrt{3}}{2}$$ , the angle is now closer to the x-axis. This angle is $$\frac{5\pi}{6}$$ .

Note: Unlike inverse sine and inverse tangent, all inverse cosine values are nonnegative. The resulting angles can only be from $$\left[0,\ \pi\right]$$ .

Only $$\cos^{-1}\left(-1\right)$$ is positive.

Both inverse sine and inverse tangent values of a negative number are negative.

$$\cos^{-1}\left(-1\right)\ =\ \pi$$

$$\sin^{-1}\left(-1\right)\ =\ -\frac{\pi}{2}$$

$$\tan^{-1}\left(-1\right)\ =\ -\frac{\pi}{4}$$

The tangent value $$-\sqrt{3}$$ can be found in Quadrant IV where the angles are negative. The only possible choices are the inverse sine values since both inverse cosine values are positive.

Answer is $$\sin^{-1}\left(-\frac{\sqrt{3}}{2}\right)$$ . This is consistent with the ratio identity $$\tan\theta\ =\ \frac{\sin\theta}{\cos\theta}$$ .