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How do you prove #cosh(2x) = cosh²x + sinh²x#?

Answer 1

Emery Adkins

see below

Use formula: #cosh x=(e^x+e^-x)/2 and sinh x = (e^x-e^-x)/2#

Right side:#=cosh^2x+sinh^2x#

#=((e^x+e^-x)/2)^2 +((e^x-e^-x)/2)^2#

#=(e^(2x)+2+e^(-2x))/4 + (e^(2x)-2+e^(-2x))/4#

#=(e^(2x)+2+e^(-2x)+e^(2x)-2+e^(-2x))/4#

#=(2e^(2x)+2e^(-2x))/4#

#=(2(e^(2x)+e^(-2x)))/4#

#=(e^(2x)+e^(-2x))/2#

#=cosh (2x)#

#=# Left Side

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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