题目内容：

函数f(x)=(1+1/tanx)sin^2x-2sin(x+π/4)sin(x-π/4).若tana=2,求f(a).若x属于[π/12,π/2],求取值范围
函数f(x)=(1+1/tanx)sin^2x-2sin(x+π/4)sin(x-π/4).若tana=2,求f(a).若x属于[π/12,π/2]求f(x)取值范围

优质解答

f(x)=sin^2x+sinxcosx-sin^2x+cos^2x
=sinxcosx+cos^2x
=sin2x/2+(1+cos2x)/2
=sin2x/2+cos2x/2+1/2
(1)f(a)=sin2a/2+cos2a/2+1/2
根据万能公式得 sin2a=tana/(1+tan^2a)=2/5 cos2a=(1-tan^2a)/(1+tan^2a)=-3/5
f(a)=1/5-3/10+1/2=2/5
(2)f(x)=sin2x/2+cos2x/2+1/2=√2sin(2x+π/4)/2+1/2
π/12

追问：

f(x)=(1+1/tanx)sin^2x-2sin(x+π/4)sin(x-π/4).怎么变这个f(x)=sin^2x+sinxcosx-sin^2x+cos^2x