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.
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.
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'.
Hello, rman144!
There is no identity for a "mixed" statement like this.
. . We must derive one like galactus did.
Write all of them in exponential form . . .$\displaystyle \cosh(x)\sin(y)-\sinh(x)\cos(y)$
$\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] $
. . $\displaystyle = \;\;-\frac{1}{2}\bigg[(1+i)\sinh(x+iy) \;+ \;(1-i)\cosh(x-iy)\bigg]$