已知f(sin x/2)=1+cos x 求f(x)
f(sin x/2)=1+cos x 求f(x)
f(sin x/2)=1+cos x =2[cos(x/2)]=2[1-sin∧(x/2)] f(x)=2(1-x∧2)
已知f(sin x/2)=1+cos x 求f(x) f(sin)=1+cosx=1+[1-2sin^2 (x/2)]=2-2sin^2 (x/2) 那么,令sin(x/2)=t,则: f(t)=2-2t^2 所以,f(x)=2-2x^2
答:解:x²-3x+1=0 x-3+1/x=0 x+1/x=3 x²+1/x²=(x+1/x)²-2=7详情>>
答:详情>>