Thread: proving hyperbolic Identities

Prove the identities:
1.)

2.)

thanks ahead

Originally Posted by ^_^Engineer_Adam^_^

Prove the identities:
1.)

2.)

thanks ahead

1.) sinh(x+y) = sinh(x)cosh(y) +cosh(x)sin(y)

I have done this before.
I don't know how link, so here is a "copy-paste" of it:

----sinh u = (1/2)[e^u -e^(-u)]
----cosh u = (1/2)[e^u +e^(-u)]
----a^b * a^c = a^(b+c)

sinh(x+y) = sinh x cosh y + cosh x sinh y

We develop the RHS.
RHS =
= {(1/2)[e^x -e^(-x)] *(1/2)[e^y +e^(-y)]} +{(1/2)[e^x +e^(-x)] *(1/2)[e^y -e^(-y)]}

= (1/4)[e^x -e^(-x)][e^y +e^(-y)] +(1/4)[e^x +e^(-x)][e^y -e^(-y)]

Doing FOILs in the expansion,

= (1/4)[e^(x+y) +e^(x-y) -e^(-x+y) -e^(-x-y)] +(1/4)[e^(x+y) -e^(x-y) +e^(-x+y) -e^(-x-y)]

= (1/4)[e^(x+y) +e^(x-y) -e^(-x+y) -e^(-x-y) +e^(x+y) -e^(x-y) +e^(-x+y) -e^(-x-y)]

= (1/4)[2e^(x+y) -2e^(-x-y)]

= (1/2)[e^(x+y) -e^(-x-y)]

= (1/2)[e^(x+y) -e^[-(x+y)]

= sinh (x+y)

= LHS

Therefore, proven.

-------------------------------------------
2) tanh(ln(x)) = (x^2 -1) /(x^2 +1)

tanh(ln(x)) =
= sinh(ln(x)) /cosh(ln(x))
= (e^ln(x) -e^(-ln(x)) / (e^ln(x) +e^(-ln(x))
= (x -x^(-1)) / (x +x^(-1))
= (x -1/x) / (x +1/x)
= [(x^2 -1)/x] / [(x^2 +1)/x]
= (x^2 -1) / (x^2 +1)

Proven.

Note:
e^(-ln(x))
= 1 /e^ln(x)
= 1 /x

Originally Posted by ticbol

1.) sinh(x+y) = sinh(x)cosh(y) +cosh(x)sin(y)

I have done this before.
I don't know how link, so here is a "copy-paste" of it:

----sinh u = (1/2)[e^u -e^(-u)]
----cosh u = (1/2)[e^u +e^(-u)]
----a^b * a^c = a^(b+c)

sinh(x+y) = sinh x cosh y + cosh x sinh y

We develop the RHS.
RHS =
= {(1/2)[e^x -e^(-x)] *(1/2)[e^y +e^(-y)]} +{(1/2)[e^x +e^(-x)] *(1/2)[e^y -e^(-y)]}

= (1/4)[e^x -e^(-x)][e^y +e^(-y)] +(1/4)[e^x +e^(-x)][e^y -e^(-y)]

Doing FOILs in the expansion,

= (1/4)[e^(x+y) +e^(x-y) -e^(-x+y) -e^(-x-y)] +(1/4)[e^(x+y) -e^(x-y) +e^(-x+y) -e^(-x-y)]

= (1/4)[e^(x+y) +e^(x-y) -e^(-x+y) -e^(-x-y) +e^(x+y) -e^(x-y) +e^(-x+y) -e^(-x-y)]

= (1/4)[2e^(x+y) -2e^(-x-y)]

= (1/2)[e^(x+y) -e^(-x-y)]

= (1/2)[e^(x+y) -e^[-(x+y)]

= sinh (x+y)

= LHS

Therefore, proven.

-------------------------------------------

but its done backwards ...
i followed from the end to the start
but im still confused how
= (1/2)[e^(x+y) -e^(-x-y)]
became
= (1/4)[2e^(x+y) -2e^(-x-y)]

Originally Posted by ^_^Engineer_Adam^_^

but its done backwards ...
i followed from the end to the start
but im still confused how
= (1/2)[e^(x+y) -e^(-x-y)]
became
= (1/4)[2e^(x+y) -2e^(-x-y)]

Because he factored out a two,
(2/4)[e^(x+y)-e^(-x-y)]
Simplify,
(1/2)[e^(x+y)-e^(-x-y)]

Hello, ^_^Engineer_Adam^_^ !

This is ticbol's solution . . . in LaTeX.
. . (Don't mind me . . . I love doing this!)


Ticbol wrote:



. .

We have: .

We develop the RHS.



. .

Doing FOILs in the expansion:

. .

. .

. .

. .

Therefore, proven.


~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~






Multiply by

Therefore, proven.