Question

设函数f(x)=sinxcosx+cos²x

Answer 1

f(x)=cos²x+sinxcosx
=1/2+1/2cos2x+1/2sin2x
=1/2+√2/2(√2/2cos2x+√2/2sin2x)
=1/2+√2/2(cosπ/4cos2x+sinπ/4sin2x)
=1/2+√2/2cos(2x-π/4)
(1)当cos(2x-π/4)=1时又ymax=1/2+√2/2
(2)将x=a/2+π/8代入f(x)中可得
f(a/2+π/8)=1/2+√2/2cos[2(a/2+π/8)-π/4]=1,解得a=π/4
所以s△abc=1/2bcsina=1/2×3×3×sinπ/4=9√2/4
希望对你有所帮助.

Answer 2

有些符号打不出来，见谅了！
f(x)=sinxcosx+cos^2
=1/2sin2x+1/2(cos2x+1)
=1/2(sin2x+cos2x)+1/2
=根号2/2sin(2x+π/4)+1/2
所以周期为π;
当x∈【0，π/2】时；
2x+π/4∈【π/4，3π/4】
则当2x+π/4=π/2时，f(x)有最大值，为
根号2/2+1/2;
当2x+π/4=3π/4时，f(x)有最小值，为
负根号2/2+1/2