How to prove the accuracy of this equation?

Question

$$(\sinh x + \cosh x)^{n} = \sinh nx + \cosh nx$$

I already know:

$$\sinh x= \frac{e^{x} - e^{-x}}{2}$$ And $$\cosh x= \frac{e^{x} + e^{-x}}{2}$$ But I failed to prove it, since I don't see any point to start from.

Use the two defitions you know and substitute them into the LHS and RHS of your equation. — Blitzer, Nov 03 '22 at 17:24
MathJax hint: if you put a backslash before common functions you get the right font ans spacing. So \sinh x gives $\sinh x$ instead of sinh x giving $sinh x$ — Ross Millikan, Nov 03 '22 at 17:26

score 4 · Accepted Answer · answered Nov 03 '22 at 18:00

Thanks to the hints of Blitzer and David Quinn found a way.

Simplified LHS: $$({\frac{e^{x} - e^{-x} + e^{x} + e^{-x}}{2}})^{n} = $$ $$(\frac{2e^{x}}{2})^{n} = e^{nx}$$

Simplified RHS: $${\frac{e^{nx} - e^{-nx} + e^{nx} + e^{-nx}}{2}} = $$ $$\frac{2e^{nx}}{2} = e^{nx}$$

LHS = RHS, so proved. Thanks.

score 1 · Answer 2 · answered Nov 15 '22 at 05:44

1

$$(\sinh (x) + \cosh (x))^{n} = \sinh (nx) + \cosh (nx)$$ Let $x=i\,y$ to make $$(\cos (y)+i \sin (y))^n=\cos (n y)+i \sin (n y)$$

answered Nov 15 '22 at 05:44

Claude Leibovici

2 Answers2