Find the values in Exercises 9–12.
11. tan(arcsin(-1/2))

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Recognize that the expression is \( \tan(\arcsin(-\frac{1}{2})) \). Here, \( \arcsin(-\frac{1}{2}) \) represents an angle \( \theta \) such that \( \sin(\theta) = -\frac{1}{2} \).

Let \( \theta = \arcsin(-\frac{1}{2}) \). Since \( \sin(\theta) = -\frac{1}{2} \), we want to find \( \tan(\theta) \). Recall that \( \tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} \).

Use the Pythagorean identity to find \( \cos(\theta) \): \( \cos(\theta) = \pm \sqrt{1 - \sin^2(\theta)} = \pm \sqrt{1 - \left(-\frac{1}{2}\right)^2} = \pm \sqrt{1 - \frac{1}{4}} = \pm \sqrt{\frac{3}{4}} = \pm \frac{\sqrt{3}}{2} \).

Determine the correct sign of \( \cos(\theta) \) by considering the range of \( \arcsin \), which is \( \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \). In this interval, cosine is positive for angles where sine is negative, so \( \cos(\theta) = \frac{\sqrt{3}}{2} \).

Finally, compute \( \tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} = \frac{-\frac{1}{2}}{\frac{\sqrt{3}}{2}} = -\frac{1}{\sqrt{3}} \).

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Inverse Trigonometric Functions

Inverse trigonometric functions, like arcsin, return the angle whose trigonometric ratio equals a given value. For example, arcsin(-1/2) gives the angle θ in the range [-π/2, π/2] such that sin(θ) = -1/2.

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To find tan(arcsin(x)), use the identity tan(θ) = sin(θ)/cos(θ). Knowing sin(θ), you can find cos(θ) using the Pythagorean identity cos²(θ) = 1 - sin²(θ), then compute the tangent.

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The Pythagorean identity states that sin²(θ) + cos²(θ) = 1 for any angle θ. This allows calculation of one trigonometric function if the other is known, essential for evaluating expressions like tan(arcsin(x)).

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