设f(x)=(e^x-e^-x)/2,g(x)=(e^x+e^-x)/2,求证:)
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设f(x)=(e^x-e^-x)/2,g(x)=(e^x+e^-x)/2,求证:)
设f(x)=[(e^x)-(e^-x)]/2,g(x)=[(e^x)+(e^-x)]/2,求证:
(1)[g(x)]^2-[f(x)]^2=1
(2)f(2x)=2f(x)·g(x),(注意“·”为乘号)
(3)g(2x)=[g(x)]^2+[f(x)]^2
设f(x)=[(e^x)-(e^-x)]/2,g(x)=[(e^x)+(e^-x)]/2,求证:
(1)[g(x)]^2-[f(x)]^2=1
(2)f(2x)=2f(x)·g(x),(注意“·”为乘号)
(3)g(2x)=[g(x)]^2+[f(x)]^2
设f(x)=[e^x-e^(-x)]/2,g(x)=[e^x+e^(-x)]/2,求证:
(1)[g(x)]^2-[f(x)]^2=1
(2)f(2x)=2f(x)×g(x)
(3)g(2x)=[g(x)]^2+[f(x)]^2
1)
[g(x)]^2-[f(x)]^2=1
[g(x)]^2-[f(x)]^2
=[e^x+e^(-x)]^2/4-[e^x-e^(-x)]^2/4
=[e^2x+e^-2x+2)-(e^2x+e^-2x-2)]/4
=4/4
=1
2)
f(2x)=(e^2x-e^-2x)/2
=(e^x+e^-x)(e^x-e^-x)/2
=2f(x)g(x)
3)
g(2x)=[g(x)]^2+[f(x)]^2
[g(x)]^2+[f(x)]^2
=[e^x+e^(-x)]^2/4+[e^x-e^(-x)]^2/4
=[e^2x+e^-2x+2)+(e^2x+e^-2x-2)]/4
=(2e^2x+2e^-2x)/4
=(e^2x+e^-2x)/2
=g(2x)
得证
(1)[g(x)]^2-[f(x)]^2=1
(2)f(2x)=2f(x)×g(x)
(3)g(2x)=[g(x)]^2+[f(x)]^2
1)
[g(x)]^2-[f(x)]^2=1
[g(x)]^2-[f(x)]^2
=[e^x+e^(-x)]^2/4-[e^x-e^(-x)]^2/4
=[e^2x+e^-2x+2)-(e^2x+e^-2x-2)]/4
=4/4
=1
2)
f(2x)=(e^2x-e^-2x)/2
=(e^x+e^-x)(e^x-e^-x)/2
=2f(x)g(x)
3)
g(2x)=[g(x)]^2+[f(x)]^2
[g(x)]^2+[f(x)]^2
=[e^x+e^(-x)]^2/4+[e^x-e^(-x)]^2/4
=[e^2x+e^-2x+2)+(e^2x+e^-2x-2)]/4
=(2e^2x+2e^-2x)/4
=(e^2x+e^-2x)/2
=g(2x)
得证
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