Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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![**Simplify the following function until there are no trigonometric expressions.**
\[ \sin(\arccos x) = \]
To simplify \(\sin(\arccos x)\), consider a right triangle where \(\theta = \arccos x\). In this triangle:
- The adjacent side to angle \(\theta\) is \(x\).
- The hypotenuse is 1 (since the cosine is adjacent over hypotenuse).
Using the Pythagorean theorem, the opposite side, which we need for sin, is \(\sqrt{1-x^2}\).
Thus, \(\sin(\arccos x) = \sqrt{1-x^2}\).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F17a10252-ae51-4d74-b5c4-8b720ae84599%2Fa88bc598-3517-44c4-b3cc-e0240b24e502%2Foxup0v_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Simplify the following function until there are no trigonometric expressions.**
\[ \sin(\arccos x) = \]
To simplify \(\sin(\arccos x)\), consider a right triangle where \(\theta = \arccos x\). In this triangle:
- The adjacent side to angle \(\theta\) is \(x\).
- The hypotenuse is 1 (since the cosine is adjacent over hypotenuse).
Using the Pythagorean theorem, the opposite side, which we need for sin, is \(\sqrt{1-x^2}\).
Thus, \(\sin(\arccos x) = \sqrt{1-x^2}\).
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