hyperbolic functions-- prove a relationship

[dts5044]

prove: sinh[sup:398ukeea]2[/sup:398ukeea](a+b) - sinh[sup:398ukeea]2[/sup:398ukeea](a-b) = sinh(2a)sinh(2b)

Another tough one I was assigned...I thought of bringing sinh(2a)sinh(2b) to the left side and setting f(x) equal to that expression, then checking out the derivative. Not sure how to get around (a+b)[sup:398ukeea]'[/sup:398ukeea] on that. Should I just try using the properties of hyperbolic functions to rearrange?

 

[markbrock30]

Sinh = ( e(x) - e(-x) / 2)

Sinh^2(x) = ( e(x) - e(-x) / 2).( e(x) - e(-x) / 2)

So you can write the left hand side of the equation as:

( e(a + b) - e(-(a +b)) / 2).( e(a + b) - e(-(a +b)) / 2) - ( e(a - b) - e(-(a - b)) / 2).( e(a - b) - e(-(a - b)) / 2)

You can prove it by re-arranging from there... But i'll leave that up too you .

Note: Your trying to ultimately get to:

( 2e(2a)) - 2(e(-2a)) + 2(e(2b)) - 2(e(-2b)) ) / 4

Which is the exponential description of the right hand side of the equation.

Mark.

 

[soroban]

Hello, dts5044!

Are we allowed to use basic hyperbolic identities?

. . \(\displaystyle \sinh(x \pm y) \;=\;\sinh(x)\cosh(y) \pm \cosh(x)\sinh(y)\)

. . \(\displaystyle \sinh(2x) \;=\;2\sinh(x)\cosh(x)\)

\(\displaystyle \text{Prove: }\;\sinh^2(a+b) - \sinh^2(a-b) \;= \;\sinh(2a)\sinh(2b)\)

\(\displaystyle \sinh^2(a+b) \:=\:\left[\sinh(a)\cosh(b) + \cosh(a)\sin(b)\right]^2\)
. . . . . . . . . . . . \(\displaystyle =\;\sinh^2(a)\cosh^2(b) + 2\sinh(a)\cosh(a)\sinh(b)\cosh(b) + \cosh^2(a)\sinh^2(b)\) . [1]

\(\displaystyle \sinh^2(a-b) \:=\:\left[\sinh(a)\cosh(b) - \cosh(a)\sinh(b)\right]^2\)
. . . . . . . . . . . . \(\displaystyle =\;\sinh^2(a)\cosh^2(b) - 2\sinh(a)\cosh(a)\sinh(b)\cosh(b) + \cosh^2(a)\sinh^2(b)\) . [2]


Subtract [1] - [2]

\(\displaystyle \sinh^2(a+b) - \sinh^2(a-b) \;=\;4\sinh(a)\cosh(a)\sinh(b)\cosh(b)\)

. . . . . . . . . . . . . . . . . . . . . . . . . \(\displaystyle = \;[2\sinh(a)\cosh(a)][2\sinh(b)\cosh(b)]\)

. . . . . . . . . . . . . . . . . . . . . . . . . \(\displaystyle = \;\sinh(2a)\sinh(2b)\)

 

[dts5044]

I would assume basic identities are valid in proving the solution, so that seems the simpler answer. Thank you both for the help!