In Exercises 39–54, find the exact value of each expression, if possible. Do not use a calculator. tan⁻¹ [tan(− π/6)]

검증된 단계별 안내

Recall that the function \( \tan^{-1}(x) \), also known as arctangent, returns an angle \( \theta \) such that \( -\frac{\pi}{2} < \theta < \frac{\pi}{2} \) and \( \tan(\theta) = x \). This means the output of \( \tan^{-1} \) is always in the principal range \( (-\frac{\pi}{2}, \frac{\pi}{2}) \).

Evaluate the inner function first: \( \tan(-\frac{\pi}{6}) \). Since tangent is an odd function, \( \tan(-x) = -\tan(x) \), so \( \tan(-\frac{\pi}{6}) = -\tan(\frac{\pi}{6}) \).

Recall the exact value \( \tan(\frac{\pi}{6}) = \frac{1}{\sqrt{3}} \), so \( \tan(-\frac{\pi}{6}) = -\frac{1}{\sqrt{3}} \).

Now substitute back into the original expression: \( \tan^{-1} \left( -\frac{1}{\sqrt{3}} \right) \). We need to find the angle \( \theta \) in the principal range \( (-\frac{\pi}{2}, \frac{\pi}{2}) \) such that \( \tan(\theta) = -\frac{1}{\sqrt{3}} \).

Since \( \tan(\theta) = -\frac{1}{\sqrt{3}} \) and \( \tan(-\frac{\pi}{6}) = -\frac{1}{\sqrt{3}} \), the angle \( \theta = -\frac{\pi}{6} \) lies within the principal range, so \( \tan^{-1} [ \tan(-\frac{\pi}{6}) ] = -\frac{\pi}{6} \).

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주요 개념

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Inverse Trigonometric Functions

Inverse trigonometric functions, like tan⁻¹ (arctan), return the angle whose trigonometric ratio equals a given value. For arctan, the output angle lies within the principal range of (−π/2, π/2). Understanding this range is crucial for correctly interpreting inverse function values.

Periodicity of the Tangent Function

The tangent function is periodic with period π, meaning tan(θ) = tan(θ + nπ) for any integer n. This property helps simplify expressions involving tangent by reducing angles to an equivalent angle within a specific interval.

Evaluating Composite Functions Involving Inverse Trigonometric Functions

When evaluating expressions like tan⁻¹[tan(θ)], the result is the angle in the principal range of arctan that is coterminal with θ modulo π. If θ lies outside this range, the value must be adjusted to find the equivalent angle within (−π/2, π/2).