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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Question
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![**Prove the Identity:**
\[
\sinh(x + y) = \sinh(x) \cosh(y) + \cosh(x) \sinh(y)
\]
\[
\sinh(x) \cosh(y) + \cosh(x) \sinh(y) = \left[\frac{1}{2}(e^x - e^{-x})\right]\left[\frac{1}{2}\left(e^y + e^{-y}\right)\right] + \left[\frac{1}{2}(e^y - e^{-y})\right]\left[\frac{1}{2}(e^x + e^{-x})\right]
\]
\[
= \frac{1}{4} \left( (e^x - e^{-x})(e^y + e^{-y}) + (e^y - e^{-y})(e^x + e^{-x}) \right)
\]
\[
= \frac{1}{4} \left( [e^{x+y} + e^{x-y} - e^{-x+y} - e^{-x-y}] + [e^{x+y} - e^{x-y} + e^{-x+y} - e^{-x-y}] \right)
\]
\[
= \frac{1}{4} \left( 2e^{x+y} - 2e^{-x-y} \right)
\]
\[
= \frac{1}{2} \left[ e^{x+y} - e^{-(x+y)} \right]
\]
\[
= \sinh(x + y)
\]
**Explanation:**
The equation demonstrates the proof of the hyperbolic identity for the sum of angles using the definitions of hyperbolic sine (\(\sinh\)) and cosine (\(\cosh\)) functions. The steps involve expanding and simplifying products and sums of exponential functions, showing that both sides of the identity are equal.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F5969d757-5ec4-415f-9475-4784dbb235a2%2F431d8f72-bce3-4da5-aacb-e51afe3a148e%2F2rdrsrk_processed.png&w=3840&q=75)
Transcribed Image Text:**Prove the Identity:**
\[
\sinh(x + y) = \sinh(x) \cosh(y) + \cosh(x) \sinh(y)
\]
\[
\sinh(x) \cosh(y) + \cosh(x) \sinh(y) = \left[\frac{1}{2}(e^x - e^{-x})\right]\left[\frac{1}{2}\left(e^y + e^{-y}\right)\right] + \left[\frac{1}{2}(e^y - e^{-y})\right]\left[\frac{1}{2}(e^x + e^{-x})\right]
\]
\[
= \frac{1}{4} \left( (e^x - e^{-x})(e^y + e^{-y}) + (e^y - e^{-y})(e^x + e^{-x}) \right)
\]
\[
= \frac{1}{4} \left( [e^{x+y} + e^{x-y} - e^{-x+y} - e^{-x-y}] + [e^{x+y} - e^{x-y} + e^{-x+y} - e^{-x-y}] \right)
\]
\[
= \frac{1}{4} \left( 2e^{x+y} - 2e^{-x-y} \right)
\]
\[
= \frac{1}{2} \left[ e^{x+y} - e^{-(x+y)} \right]
\]
\[
= \sinh(x + y)
\]
**Explanation:**
The equation demonstrates the proof of the hyperbolic identity for the sum of angles using the definitions of hyperbolic sine (\(\sinh\)) and cosine (\(\cosh\)) functions. The steps involve expanding and simplifying products and sums of exponential functions, showing that both sides of the identity are equal.
Expert Solution
Step 1
Given identity for proof is
Sinh(x+y) = Sinh(x)Cosh(y)+ Cosh(x)Sinh(y)
Step by step
Solved in 2 steps with 1 images
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