Question

Prove that `(coshx+sinhx)^n=coshnx+sinhnx`?

Answer 1

Please see below.

Explanation:

As `coshx=(e^x+e^(-x))/2` and `sinhx=(e^x-e^(-x))/2`

and adding two `coshx+sinhx=e^x`

and `(coshx+sinhx)^n=(e^x)^n=e^(nx)`

also `coshnx=(e^(nx)+e^(-nx))/2` and `sinhnx=(e^(nx)-e^(-nx))/2`

and `coshnx+sinhnx=e^(nx)`

Hence, `(coshx+sinhx)^n=coshnx+sinhnx`

Answer 2

Please see the proof below

Explanation:

`" Proof by mathematical induction "`

Let the property be `P(n)`

`(coshx+sinhx)^n=cosh(nx)+sinh(nx)`

For `P(1)`

`n=1`, `=>`, `(coshx+sinhx)^1=cosh(1x)+sinh(1x)`

This is true for `n=1`

Assume that this is true for `P(n)`

`(coshx+sinhx)^n=cosh(nx)+sinh(nx)`

Then, for `P(n+1)`

`(coshx+sinhx)^(n+1)=(coshx+sinhx)^n(coshx+sinhx)`

`=(cosh(nx)+sinh(nx))(coshx+sinhx)`

`=coshnxcoshx+sinhnxsinhx+sinhnxcoshx+coshnxsinhx`

`=cosh(nx+x)+sinh(nx+x)`

`=cosh(n+1)x+sinh(n+1)x`

This is true for `P(n+1)`

As this is true for `n=1` to `n=n+1`, this is true for all `n`