1. ## proving hyperbolic Identities

Prove the identities:
1.) $\sinh(x+y) = \sinh{x}\cosh{y} + \cosh{x}\sinh{y}$

2.) $\tanh(\ln{x}) = \frac{x^2 - 1}{x^2 + 1}$

Prove the identities:
1.) $\sinh(x+y) = \sinh{x}\cosh{y} + \cosh{x}\sinh{y}$

2.) $\tanh(\ln{x}) = \frac{x^2 - 1}{x^2 + 1}$

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

3. 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)]

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)]

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

Ticbol wrote:

$1)\;\sinh(x+y) \:= \:\sinh(x)\cosh(y) + \cosh(x)\sin(y)$

. . $\sinh u \,= \,\frac{e^u -e^{-u}}{2}\qquad \cosh u \,= \,\frac{e^u + e^{-u}}{2} \qquad (a^b)(a^c)\,=\,a^{b+c}$

We have: . $\sinh(x+y) \:= \:\sinh x \cosh y + \cosh x \sinh y$

We develop the RHS.

$\text{RHS }\; = \;\left(\frac{e^x -e^{-x}}{2}\right)\left(\frac{e^y +e^{-y}}{2}\right) + \left(\frac{e^x +e^{-x}}{2}\right)\left(\frac{e^y -e^{-y}}{2}\right)$

. . $= \;\frac{(e^x -e^{-x})(e^y +e^{-y})}{4} + \frac{(e^x +e^{-x})(e^y -e^{-y})}{4}$

Doing FOILs in the expansion:

. . $= \;\frac{1}{4}\left[e^{x+y} +e^{x-y} -e^{-x+y} -e^{-x-y}\right] +\frac{1}{4}\left[e^{x+y} -e^{x-y} +e^{-x+y} -e^{-x-y}\right]$

. . $= \;\frac{1}{4}\left[e^{x+y} +e^{x-y} -e^{-x+y} -e^{-x-y} +e^{x+y} -e^{x-y} +e^{-x+y} -e^{-x-y}\right]$

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

. . $= \;\sinh (x+y) \;= \;\text{LHS}$

Therefore, proven.

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

$2)\;\;\tanh(\ln x ) \;= \;\frac{x^2 -1}{x^2 +1}$

$\tanh(\ln x)\;= \;\frac{\sinh(\ln x)}{\cosh(\ln x )}\;= \;\frac{\frac{e^{\ln x} -e^{-\ln x}}{2}} {\frac{e^{\ln x} + e^{-\ln x}}{2}} \;= \; \frac{e^{\ln x} -e^{-\ln x}}{e^{\ln x} + e^{-\ln x}}$ $= \; \frac{e^{\ln x} -e^{\ln x^{-1}}}{e^{\ln x} + e^{\ln x^{-1}}} \;= \;\frac{x -x^{-1}}{x +x^{-1}}$

Multiply by $\frac{x}{x}\!:\;\;\frac{x^2 - 1}{x^2 + 1}$

Therefore, proven.

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# proof of identities tanhx in e^2

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