求解满足f'=g,g'=f,f(0)=1,g(0)=0的f,g并证明指定等式

First, use the algebraic identity ( (a + b)(a - b) = a^2 - b^2 ). Applying this to our expression:
( [f(x) + g(x)][f(x) - g(x)] = f(x)^2 - g(x)^2 )

Compute ( f(x)^2 ) and ( g(x)^2 ) using their explicit forms:
( f(x)^2 = \left( \frac{e^x + e^{-x}}{2} \right)^2 = \frac{e^{2x} + 2 e^x e^{-x} + e^{-2x}}{4} = \frac{e^{2x} + 2 + e^{-2x}}{4} )
( g(x)^2 = \left( \frac{e^x - e^{-x}}{2} \right)^2 = \frac{e^{2x} - 2 e^x e^{-x} + e^{-2x}}{4} = \frac{e^{2x} - 2 + e^{-2x}}{4} )

Subtract ( g(x)^2 ) from ( f(x)^2 ):
( f(x)^2 - g(x)^2 = \frac{(e^{2x} + 2 + e^{-2x}) - (e^{2x} - 2 + e^{-2x})}{4} = \frac{4}{4} = 1 )

This identity holds for any ( x ), so substituting ( x = 10 ) gives:
( [f(10) + g(10)] \cdot [f(10) - g(10)] = 1 )

Alternatively, compute ( f(x) + g(x) ) and ( f(x) - g(x) ) directly:
( f(x) + g(x) = \frac{e^x + e^{-x}}{2} + \frac{e^x - e^{-x}}{2} = e^x )
( f(x) - g(x) = \frac{e^x + e^{-x}}{2} - \frac{e^x - e^{-x}}{2} = e^{-x} )
Multiplying these gives ( e^x \cdot e^{-x} = e^{0} = 1 ), confirming the result.

内容的提问来源于stack exchange，提问作者user143993