Prove this hyperbolic function?

Question

Cosh(A+B) = CoshA*CoshB+SinhB*SinhA

Please help

Answer 1

Cosh(A+B) = CoshA*CoshB+SinhB*SinhA

We know that Cosh (A+B) = [e^(A+B) + e^-(A+B)]/2
and
CoshA*CoshB+SinhB*SinhA =
{[(e^A) + (e^-A)]/2} {[(e^B) + (e^-B)]/2} + {[(e^A) - (e^-A)]/2} {[(e^B) - (e^-B)]/2}
then just solve the equation
{[e^(A+B)+ e^(A-B)+ e^-(A-B)+ e^-(A+B)]/4} +
{[e^(A+B)- e^(A-B)- e^-(A-B)+ e^-(A+B)]/4}
then add everything together and you'll get
[2e^(A+B) + 2e^-(A+B)]/4
simplify and you'll get
[e^(A+B) + e^-(A+B)]/2
which is what we had for the equation of the Cosh (A+B) at the beginning.

Answer 2

Easy, always go back to the definition if you ever get on something like this.

By the way, this is a hyperbolic identity, not a function so to speak,

I would start from the left side and then simplify it to the right.

The definition of cosh(x)=(e^x+e^-x)/2.
The definition of sinsh(x)=(e^x-e^-x)/2.

Plug a and b in to the above definition. Multiply and add them. A whole bunch of stuff wil cancel and you will end up with
(e^(a+b)+e^-(a+b))/2 = cosh(a+b)

I am sure that you are smart enough to figure out the things in between.

Answer 3

you need to plug every cosh(x) and sinh(X) into the e^x based definitions...then you will see the sides come out equal. tell me if you still don't understand

Answer 4

Cosh(a)Cosh(b)=
((e^a+e^-a)/2)((e^b+e^-b)/2)=
(e^(a+b)+e^(a-b)+e^(b-a)+e^-(a+b))/4

Sinh(a)Sinh(b)=
((e^a-e^-a)/2)((e^b-e^-b)/2)=
(e^(a+b)-e^(a-b)-e^(b-a)+e^-(a+b))/4

Cosh(a)Cosh(b)+Sinh(a)Sinh(b)=
(2e^(a+b)+2e^-(a+b))/4=
(e^(a+b)+e^-(a+b))/2=
Cosh(a+b)

the ... are (a+b)