Textbook Question
In Exercises 29–51, find the exact value of each expression. Do not use a calculator. sin(cos⁻¹ 3/5)
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5. Inverse Trigonometric Functions and Basic Trigonometric Equations
Inverse Sine, Cosine, & Tangent
2:44 minutesProblem 49
Textbook Question
In Exercises 39–54, find the exact value of each expression, if possible. Do not use a calculator. tan⁻¹ (tan 2π/3)
Verified step by step guidance1
Recall that the function \( \tan^{-1}(x) \), also known as the arctangent, returns an angle \( \theta \) such that \( -\frac{\pi}{2} < \theta < \frac{\pi}{2} \). This is the principal value range for the inverse tangent function.
Identify the given expression: \( \tan^{-1}(\tan(\frac{2\pi}{3})) \). We want to find the angle in the principal range whose tangent is equal to \( \tan(\frac{2\pi}{3}) \).
Calculate or recall the value of \( \tan(\frac{2\pi}{3}) \). Since \( \frac{2\pi}{3} = 120^\circ \), which lies in the second quadrant, use the tangent identity: \( \tan(\pi - x) = -\tan(x) \). So, \( \tan(\frac{2\pi}{3}) = \tan(\pi - \frac{\pi}{3}) = -\tan(\frac{\pi}{3}) \).
Evaluate \( \tan(\frac{\pi}{3}) \), which is \( \sqrt{3} \), so \( \tan(\frac{2\pi}{3}) = -\sqrt{3} \).
Now, find \( \theta = \tan^{-1}(-\sqrt{3}) \) where \( \theta \) is in the interval \( (-\frac{\pi}{2}, \frac{\pi}{2}) \). Determine the angle \( \theta \) whose tangent is \( -\sqrt{3} \) within this principal range.
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Video duration:
2mKey Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Inverse Tangent Function (tan⁻¹ or arctan)
The inverse tangent function, arctan, returns the angle whose tangent is a given number. Its output range is limited to (-π/2, π/2), meaning it only returns angles in the first and fourth quadrants. Understanding this range is crucial when simplifying expressions involving arctan.
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Inverse Tangent
Periodicity and Properties of the Tangent Function
The tangent function has a period of π, so tan(θ) = tan(θ + nπ) for any integer n. This periodicity allows angles outside the principal range to be related to angles within it, which is essential when simplifying expressions like tan⁻¹(tan(2π/3)).
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Simplifying Expressions Involving Inverse Trigonometric Functions
When evaluating expressions like tan⁻¹(tan(θ)), the result is the angle equivalent to θ but restricted to the principal range of arctan. This often requires adjusting θ by adding or subtracting multiples of π to find an equivalent angle within (-π/2, π/2).
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