# Thread: Can't Find The Identity

1. ## Can't Find The Identity

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.

2. 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'.

3. 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

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?

5. Here is one, for what it's worth:

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

6. ## Thanks

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

7. I derived it.

8. Hello, rman144!

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

$\displaystyle \cosh(x)\sin(y)-\sinh(x)\cos(y)$
Write all of them in exponential form . . .

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

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

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

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

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

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

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

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

9. ## Thanks

You guys are life savers.