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Ch. 2 - Graphs of the Trigonometric Functions; Inverse Trigonometric Functions
Blitzer - Trigonometry 3rd Edition
Blitzer3rd EditionTrigonometryISBN: 9780137316601Not the one you use?Change textbook
Chapter 2, Problem 87

In Exercises 83–94, use a right triangle to write each expression as an algebraic expression. Assume that x is positive and that the given inverse trigonometric function is defined for the expression in x. cos (sin⁻¹ 1/x)

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Recognize that the expression is \( \cos(\sin^{-1}(1/x)) \). Let \( \theta = \sin^{-1}(1/x) \), which means \( \sin(\theta) = \frac{1}{x} \).
Since \( \sin(\theta) = \frac{\text{opposite}}{\text{hypotenuse}} \), represent this as a right triangle where the opposite side is 1 and the hypotenuse is \( x \).
Use the Pythagorean theorem to find the adjacent side of the triangle: \( \text{adjacent} = \sqrt{x^2 - 1^2} = \sqrt{x^2 - 1} \).
Recall that \( \cos(\theta) = \frac{\text{adjacent}}{\text{hypotenuse}} \). Substitute the values from the triangle to get \( \cos(\sin^{-1}(1/x)) = \frac{\sqrt{x^2 - 1}}{x} \).
Write the final algebraic expression for the original expression as \( \frac{\sqrt{x^2 - 1}}{x} \), assuming \( x > 0 \) to keep the square root defined and positive.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Inverse Trigonometric Functions

Inverse trigonometric functions, like sin⁻¹(x), return the angle whose trigonometric ratio equals x. Understanding that sin⁻¹(1/x) gives an angle θ such that sin(θ) = 1/x is essential for rewriting expressions involving these functions.
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Introduction to Inverse Trig Functions

Right Triangle Definitions of Trigonometric Ratios

Trigonometric ratios (sine, cosine, tangent) can be represented as ratios of sides in a right triangle. By constructing a right triangle with an angle θ where sin(θ) = opposite/hypotenuse = 1/x, we can find other ratios like cos(θ) in terms of x.
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Solving Right Triangles with the Pythagorean Theorem

Pythagorean Theorem

The Pythagorean theorem relates the sides of a right triangle: (hypotenuse)² = (opposite)² + (adjacent)². This allows us to find the missing side length when two sides are known, which is crucial for expressing cos(sin⁻¹(1/x)) algebraically.
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Solving Right Triangles with the Pythagorean Theorem
Related Practice
Textbook Question

In Exercises 83–94, use a right triangle to write each expression as an algebraic expression. Assume that x is positive and that the given inverse trigonometric function is defined for the expression in x. sec (cos⁻¹ 1/x)

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Textbook Question

In Exercises 83–94, use a right triangle to write each expression as an algebraic expression. Assume that x is positive and that the given inverse trigonometric function is defined for the expression in x. csc (cot⁻¹ x)

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In Exercises 83–94, use a right triangle to write each expression as an algebraic expression. Assume that x is positive and that the given inverse trigonometric function is defined for the expression in x. sin (tan⁻¹ x)

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In Exercises 79–82, graph f, g, and h in the same rectangular coordinate system for 0 ≤ x ≤ 2π. Obtain the graph of h by adding or subtracting the corresponding y-coordinates on the graphs of f and g. f(x) = cos x, g(x) = sin 2x, h(x) = (f − g)(x)

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Textbook Question

In Exercises 83–94, use a right triangle to write each expression as an algebraic expression. Assume that x is positive and that the given inverse trigonometric function is defined for the expression in x. ___ sec (sin⁻¹ x/√x²+4)

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Textbook Question

In Exercises 83–94, use a right triangle to write each expression as an algebraic expression. Assume that x is positive and that the given inverse trigonometric function is defined for the expression in x. tan (cos⁻¹ x)

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