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

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

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

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

$\bold{\implies \frac{1}{e^{nx}}}$

RHS = cosh(nx) - sinh(nx)

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

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

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

$\bold{\implies \frac{1}{e^{nx}}}$

hence, LHS = RHS

I HOPE IT'S HELP YOU DEAR,
THANKS