原题:若f(tanx)=sinxcosx,则f(2/3)的值
相当于已知tanx=2/3,求sinxcosx
则 sinxcosx
=sinxcosx/(sin²x+cos²x)
分子分母同时除以cos²x
=tanx/(tan²x+1)
=(2/3)/[(4/9)+1]
=(2/3)/(13/9)
=6/13
原题:若f(tanx)=sinxcosx,则f(2/3)的值
相当于已知tanx=2/3,求sinxcosx
则 sinxcosx
=sinxcosx/(sin²x+cos²x)
分子分母同时除以cos²x
=tanx/(tan²x+1)
=(2/3)/[(4/9)+1]
=(2/3)/(13/9)
=6/13