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.
热心网友
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。因为f(x)+g(x)=e^x , g(x)-f(x)= 1/e^x所以[g(x)]^2-[f(x)]^2=[f(x)+g(x)]*[g(x)-f(x)]= e^x * 1/e^x = 1f(2x)= (1/2)*[e^(2x)- 1/e^(2x)]2*f(x)*g(x)= 2*(1/4)*(e^x - 1/e^x)(e^x + 1/e^x)= (1/2)*[e^(2x)- 1/e^(2x)]所以f(2x)=2f(x)*g(x)g(2x)= (1/2)*[e^(2x) + 1/ e^(2x)][g(x)]^2 + [f(x)]^2=(1/4)*[e^x+2+ 1/e^x+e^x -2+ 1/e^x] =(1/2)*[e^(2x) + 1/ e^(2x)]所以g(2x)=[g(x)]^2+[f(x)]^2 。
热心网友
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。因为f(x)+g(x)=e^x , g(x)-f(x)= 1/e^x所以[g(x)]^2-[f(x)]^2=[f(x)+g(x)]*[g(x)-f(x)]= e^x * 1/e^x = 1f(2x)= (1/2)*[e^(2x)- 1/e^(2x)]2*f(x)*g(x)= 2*(1/4)*(e^x - 1/e^x)(e^x + 1/e^x)= (1/2)*[e^(2x)- 1/e^(2x)]所以f(2x)=2f(x)*g(x)g(2x)= (1/2)*[e^(2x) + 1/ e^(2x)][g(x)]^2 + [f(x)]^2=(1/4)*[e^x+2+ 1/e^x+e^x -2+ 1/e^x] =(1/2)*[e^(2x) + 1/ e^(2x)]所以g(2x)=[g(x)]^2+[f(x)]^2。