Question

Prove the following identities.
$\sin h(x-y)=\sin hx\cos hy-\cos hx\sin hy$

Answer 1

We know that;

$coshx=\frac{e^x+e^{-x}}{2}$

$sinhx=\frac{e^x-e^{-x}}{2}$

$\to sinhxcoshy=\frac{1}{2}(e^x-e^{-x})\times \frac{1}{2}(e^y+e^{-y})$$=\frac{1}{4}(e^{x+y}+e^{x-y}-e^{-(x-y)}-e^{-(x+y)})$

$\to coshxsinhy=\frac{1}{2}(e^x+e^{-x})\times \frac{1}{2}(e^y-e^{-y})$$=\frac{1}{4}(e^{x+y}-e^{x-y}+e^{-(x-y)}-e^{-(x+y)})$

Subtracting gives:

$\to sinhxcoshy-coshxsinhy=2\times \frac{1}{4}(e^{(x-y)}-e^{-(x-y)})$$=\frac{1}{2}(e^{(x-y)}-e^{-(x-y)})=sinh(x-y)$

Hence;

$sinh(x-y)=sinhxcoshy-coshxsinhy$