设f=[sin(2/x)]=1+cosx,求f(x),f[cos(2/x)].
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设f=[sin(2/x)]=1+cosx,求f(x),f[cos(2/x)].
如题!
如题!
![设f=[sin(2/x)]=1+cosx,求f(x),f[cos(2/x)].](/uploads/image/z/6413976-0-6.jpg?t=%E8%AE%BEf%3D%5Bsin%282%2Fx%29%5D%3D1%2Bcosx%2C%E6%B1%82f%28x%29%2Cf%5Bcos%282%2Fx%29%5D.)
cosx=1-2(sinx/2)^2
f=[sin(2/x)]=1+cosx=2-2(sinx/2)^2
f(x)=2-2x^2
f[cos(2/x)]=2-2[cos(2/x)]^2
f=[sin(2/x)]=1+cosx=2-2(sinx/2)^2
f(x)=2-2x^2
f[cos(2/x)]=2-2[cos(2/x)]^2
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