Hyperbolic functions of zero

Knowing the definitions that we covered of sinh, cosh and tanh, we’re able to work out some exact values of the hyperbolic functions - by substituting for x.

\sinh0=0

Proof:

\sinh x=\frac{e^x-e^{-x}}2
\sinh0=\frac{e^0-e^{-0}}2
=\frac{1-1}2 (anything to the power of 0 is 1)
=\frac02
=0

\cosh0=1

Proof:

\cosh x=\frac{e^x+e^{-x}}2
\cosh0=\frac{e^0+e^{-0}}2
=\frac{1+1}2 (anything to the power of 0 is 1)
=\frac22
=1

\tanh0=

Proof:

\tanh x=\frac{e^x-e^{-x}}{e^x+e^{-x}}
\tanh 0=\frac{e^0-e^{-0}}{e^0+e^{-0}}
=\frac{1-1}{1+1} (anything to the power of 0 is 1)
=\frac02
=0

flashcards

Question	Answer
sinh 0	0
cosh 0	1
tanh 0	0
What is the value of \sinh 0?	0
What is the value of \cosh 0?	1
What is the value of \tanh 0?	0
How do you prove \sinh 0 = 0 using the definition?	\sinh x = \frac{e^x - e^{-x}}{2}; substitute x=0 to get \frac{e^0 - e^{-0}}{2} = \frac{1-1}{2} = 0
How do you prove \cosh 0 = 1 using the definition?	\cosh x = \frac{e^x + e^{-x}}{2}; substitute x=0 to get \frac{e^0 + e^{-0}}{2} = \frac{1+1}{2} = 1
How do you prove \tanh 0 = 0 using the definition?	\tanh x = \frac{e^x - e^{-x}}{e^x + e^{-x}}; substitute x=0 to get \frac{e^0 - e^{-0}}{e^0 + e^{-0}} = \frac{1-1}{1+1} = 0

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Outlinks

tanh cosh flashcards sinh power of 0