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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 63

In Exercises 63–82, use a sketch to find the exact value of each expression. cos (sin⁻¹ 4/5)

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Recognize that the expression is \( \cos(\sin^{-1}(\frac{4}{5})) \). Here, \( \sin^{-1}(\frac{4}{5}) \) represents an angle \( \theta \) whose sine is \( \frac{4}{5} \). So, let \( \theta = \sin^{-1}(\frac{4}{5}) \), which means \( \sin \theta = \frac{4}{5} \).
Draw a right triangle to represent the angle \( \theta \). Since \( \sin \theta = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{4}{5} \), label the side opposite to \( \theta \) as 4 and the hypotenuse as 5.
Use the Pythagorean theorem to find the adjacent side of the triangle. The formula is \( \text{adjacent} = \sqrt{\text{hypotenuse}^2 - \text{opposite}^2} = \sqrt{5^2 - 4^2} \).
Calculate the adjacent side length (do not simplify fully here, just set up the expression). This gives \( \sqrt{25 - 16} = \sqrt{9} \).
Now, find \( \cos \theta \) using the triangle sides: \( \cos \theta = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{\sqrt{9}}{5} \). This expression represents the exact value of \( \cos(\sin^{-1}(\frac{4}{5})) \).

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