Prove $\forall x \in \mathbb{R}$: $[\sinh(x)+\cosh(x)]^n = \cosh(nx)+\sinh(nx)$ ; $ n\in \mathbb{Q}$

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Prove $\forall x \in \mathbb{R}$: $[\sinh(x)+\cosh(x)]^n = \cosh(nx)+\sinh(nx)$ ; $ n\in \mathbb{Q}$

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Solution: note that $$ \sinh(x) + \cosh(x) = \frac {\mathrm e^x - \mathrm e^{-x}} 2 + \frac {\mathrm e^x + \mathrm e^{-x}} 2 = e^x, $$ and rest is obvious.