Question

y=2sinx/2,x属于[-π,0]的反函数 y=cosx x∈[-π,0] 求反函数

Answer 1

y=2sinx/2(-π<=x<0),则-2<=y<=0

sinx/2=y/2

x/2=arcsiny/2

x=2arcsiny/2

则反函数为y=2arcsinx/2(-2<=x<=0)

y=cosx x∈[-π,0],则0<=π+x<=π、-1<=y<=1

y=cosx=-cos(π+x)

cos(π+x)=-y

π+x=arccos(-y)=arccosy

x=arccos(-y)-π

反函数为y=arccos(-x)-π(-1<=x<=1)

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