Question

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

Answer 1

We know that;

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

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

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

Hence;

$cosh(x+y)=coshxcoshy+sinhxsinhy$