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Ch. 2 - Graphs of the Trigonometric Functions; Inverse Trigonometric Functions
Blitzer - Trigonometry 3rd Edition
Blitzer3rd EditionTrigonometryISBN: 9780137316601Non è quello che usi tu?Cambia libro di testo
Capitolo 2, Problema 91

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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Identify the angle \( \theta \) such that \( \theta = \sin^{-1} \left( \frac{x}{\sqrt{x^2 + 4}} \right) \). This means \( \sin \theta = \frac{x}{\sqrt{x^2 + 4}} \).
Draw a right triangle where the angle \( \theta \) has an opposite side of length \( x \) and a hypotenuse of length \( \sqrt{x^2 + 4} \) based on the sine definition \( \sin \theta = \frac{\text{opposite}}{\text{hypotenuse}} \).
Use the Pythagorean theorem to find the adjacent side of the triangle: \( \text{adjacent} = \sqrt{(\text{hypotenuse})^2 - (\text{opposite})^2} = \sqrt{(\sqrt{x^2 + 4})^2 - x^2} \).
Simplify the adjacent side expression: \( \sqrt{x^2 + 4 - x^2} = \sqrt{4} = 2 \).
Recall that \( \sec \theta = \frac{\text{hypotenuse}}{\text{adjacent}} \). Substitute the known values to write \( \sec (\sin^{-1} (\frac{x}{\sqrt{x^2 + 4}})) = \frac{\sqrt{x^2 + 4}}{2} \) as the algebraic expression.

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Inverse Trigonometric Functions

Inverse trigonometric functions, like sin⁻¹ (arcsin), return the angle whose trigonometric ratio equals a given value. Understanding their domain and range is essential, as they help translate between angle measures and ratio values, especially when constructing right triangles.
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Right Triangle Trigonometry

Right triangle trigonometry relates the sides and angles of a right triangle using ratios such as sine, cosine, and secant. By interpreting inverse trig expressions as angles in a triangle, one can find unknown sides or other trigonometric values algebraically.
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Converting trigonometric expressions involving inverse functions into algebraic forms requires substituting sides of a triangle and simplifying. This often involves using the Pythagorean theorem and expressing trigonometric ratios in terms of the variable x.
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