Can't Find The Identity

**rman144** · June 28th 2008, 12:46 PM

Anyone know of an identity for:

cosh(x)sin(y)-sinh(x)cos(y)

I know how this works if the hyperbolic cosine and sine were regular cosine and sine, but I cannot seem to find any identities online that relate these.

Thanks in advance for the help.

**bleesdan** · June 28th 2008, 12:55 PM

Hyperbolic Functions -- from Wolfram MathWorld

This has all the identities on it.

It looks like an imaginary identity, but there's no i.

Hold on

OK, I got nothin'.

**topsquark** · June 28th 2008, 01:12 PM

Originally Posted by rman144

Anyone know of an identity for:

cosh(x)sin(y)-sinh(x)cos(y)

I know how this works if the hyperbolic cosine and sine were regular cosine and sine, but I cannot seem to find any identities online that relate these.

Thanks in advance for the help.

I'm with bleesdan. I'm suspecting a typo?

-Dan

**rman144** · June 28th 2008, 01:20 PM

Sorry guys, no typo. There is no imaginary part there. This is why the problem is proving to be such an issue. Anyone else have an idea?

**galactus** · June 28th 2008, 01:32 PM

Here is one, for what it's worth:

$\frac{\sqrt{2}\left(sin(y+\frac{\pi}{4})-e^{2x}cos(y+\frac{\pi}{4})\right)}{2e^{x}}$

**rman144** · June 28th 2008, 01:37 PM

Thanks a lot. May I ask where you found that/ solved for that?

**galactus** · June 28th 2008, 01:57 PM

I derived it.

**Soroban** · June 28th 2008, 06:30 PM

Hello, rman144!

There is no identity for a "mixed" statement like this.
. . We must derive one like galactus did.

$\cosh(x)\sin(y)-\sinh(x)\cos(y)$

Write all of them in exponential form . . .

$\frac{e^x+e^{-x}}{2}\cdot\frac{e^{iy} - e^{-iy}}{2i} \;- \;\frac{e^x-e^{-x}}{2}\cdot\frac{e^{iy} + e^{-iy}}{2}$

. . $= \;\;\frac{e^{x+iy} - e^{x-iy} + e^{-x+iy} - e^{-x-iy}}{4i} \;- \;\frac{e^{x+iy} + e^{x-iy} - e^{-x+iy} - e^{-x-iy}}{4}$

Multiply the first fraction by $\frac{-i}{-i}$

$\frac{-ie^{x+iy} + ie^{x-iy} - ie^{-x+iy} + ie^{-x-iy}}{4} \;- \;\frac{e^{x+iy} + e^{x-iy} - e^{-x+iy} - e^{-x-iy}}{4}$

. . $= \;\;\frac{1}{4}\bigg[-e^{x+iy} - ie^{x+iy} - e^{x-iy} + ie^{x-iy} + e^{-x+iy} - ie^{-x+iy} + e^{-x-iy} + ie^{-x-iy}\bigg]$

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

. . $=\;\;-\frac{1}{2}\bigg[(1+i)\frac{e^{x+iy} - e^{-(x+iy)}}{2} \;+ \;(1-i)\frac{e^{x-iy} + e^{-(x-iy)}} {2}\bigg]$

. . $= \;\;-\frac{1}{2}\bigg[(1+i)\sinh(x+iy) \;+ \;(1-i)\cosh(x-iy)\bigg]$

**rman144** · June 28th 2008, 06:38 PM

You guys are life savers.

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