Question

Prove that (cosh x + sinh x)ⁿ = cosh(nx) + sinh(nx), for any n ∈ R.

Answer 1

we know,

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

LHS = $(coshx+sinhx)^n$

$=\left[\frac{e^x+e^{-x}}{2}+\frac{e^x-e^{-x}}{2}\right]^n\\\\=\left[\frac{e^x+e^{-x}+e^{x}-e^{-x}}{2}\right]^n\\\\=\left[\frac{2e^x}{2}\right]^n\\\\=e^{nx}$

now, RHS = $cosh(nx)+sinh(nx)$

$=\left[\frac{e^{nx}+e^{-nx}}{2}+\frac{e^{nx}-e^{-nx}}{2}\right]\\\\=\left[\frac{e^{nx}+e^{-nx}+e^{nx}-e^{-nx}}{2}\right]\\\\=\left[\frac{2e^{nx}}{2}\right]\\\\=e^{nx}$

hence, LHS = RHS

Answer 2

HELLO DEAR,

we know,
$\bold{coshx = \frac{e^x + e^{-x}}{2} \;\; , sinhx = \frac{e^{x} - e^{-x}}{2}}$

now,

LHS = (coshx + sinhx)^n

$\bold{\implies[\frac{e^x + e^{-x}}{2} + \frac{e^x - e^{-x}}{2}]^n}$

$\bold{\implies [\frac{e^x + e^{-x} + e^{x} - e^{-x}}{2}]^n}$

$\bold{\implies \frac{2e^{nx}}{2}}$

$\bold{\implies e^{nx}}$

RHS = cosh(nx) + sinh(nx)

$\bold{\implies [\frac{e^{nx} + e^{-nx}}{2} + \frac{e^{nx} - e^{-nx}}{2}]}$

$\bold{\implies [\frac{e^{nx} + e^{-nx} + e^{nx} + e^{-nx}}{2}]}$

$\bold{\implies [\frac{2e^{nx}}{2}]}$

$\bold{\implies e^{nx}}$

hence, LHS = RHS

I HOPE IT'S HELP YOU DEAR,
THANKS