f(x)=sinxcosx+sin²x
=1/2sin2x+(1-cos2x)/2
=1/2sin2x-1/2cos2x+1/2
=√2/2*(√2/2sin2x-√2/2cos2x)+1/2
=√2/2*sin(2x-π/4)+1/2
f(0)=√2/2*sin(2*0-π/4)+1/2
=√2/2*sin(-π/4)+1/2
=-√2/2*sinπ/4+1/2
=-√2/2*√2/2+1/2
=-1/2+1/2
=0
f(π/2)=√2/2*sin(2*π/2-π/4)+1/2
=√2/2*sin(π-π/4)+1/2
=√2/2*sinπ/4+1/2
=√2/2*√2/2+1/2
=1/2+1/2
=1sin²x为什么等于(1-cos2x)/2，这个公式我一直不明白怎么用cos2x又等于什么cos2x=2cos²x-1=1-2sin²x=cos²x-sin²x