已知函数f(x)=cosx-cos(x+π/2) x∈R.求sin2x的值

解；f(x)=cosx-cos(x+π/2)
=cosx+sinx
=√2（sinπ/4cosx+sinxcosπ/4)
=√2sin(π/4+x)
所以f(sin2x)=√2sin(π/4+sin2x)
=√2sin(π/4+2sinxcosx)
题目是否有问题?不好意思，刚才打少一个条件，已知函数f(x)=cosx-cos(x+π/2) x∈R.若f（x）＝3分之4.求sin2x的值f(x)=cosx-cos(x+π/2)=cosx+sinx=4/3所以cos²x2+2cosxsinx+sin²x=16/9所以2sinxcosx=16/9-1=7/9即sin2x=2sinxcosx=7/9