Problem

Prove the identity. $ \cosh (x + y) = \cosh x \c…

00:50

Problem 11 Easy Difficulty

Prove the identity.
$ \sinh (x + y ) = \sinh x \cosh y + \cosh x \sinh

Answer

$\frac{1}{4}\left(2 e^{x+y}-2 e^{-x-y}\right)=\frac{1}{2}\left[e^{x+y}-e^{-(x+y)}\right]=\sinh (x+y)$

Related Courses

Calculus 1 / AB

Calculus: Early Transcendentals

Chapter 3

Differentiation Rules

Section 11

Hyperbolic Functions

Discussion

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Alex O.

June 21, 2020

amrita is the only tutor on here who goes way too fast,,always skips like 20 steps

James L.

October 21, 2021

I agree. definitely shouldn't be labeled as an "educator," she explains it like everyone already knows the answer.

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Video Transcript

okay. We first started out with either the excuse y minus even negative X plus y looked closely at where the parentheses are. Gives us you did the X minus you the negative acts over too. Times eat of the Y, plus even negative. Why over too? Plus, each of the AKs puts even negative acts over too times either the y minus e to the negative. Why over to So as you can see, the night gives and the positives are alternating when we're doing this copulation. And this friendly gives us sign a jukebox co sign a chew of Why was co sign h of X sign each of why.

Amrita B.

Related Courses

Calculus 1 / AB

Calculus: Early Transcendentals

Chapter 3

Differentiation Rules

Section 11

Hyperbolic Functions

Problem

Prove the identity. $ \cosh (x + y) = \cosh x \c…

Problem 11 Easy Difficulty

Prove the identity.
$ \sinh (x + y ) = \sinh x \cosh y + \cosh x \sinh

Answer

$\frac{1}{4}\left(2 e^{x+y}-2 e^{-x-y}\right)=\frac{1}{2}\left[e^{x+y}-e^{-(x+y)}\right]=\sinh (x+y)$

Related Courses

Calculus: Early Transcendentals

Related Topics

Discussion

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