Question

(new！)高一数学 三角函数增减性问题 急求解答！！加分感谢！！

已知函数f(x)=[2sin^2(x+π/4)]-√3·cos2x（x∈[π/4,π/2]），求f(x)的最大值、最小值.
【请写出详细过程，注:答案不重要，主要是把f(x)化成Asin(ax+φ)或Acos(ax+φ)的形式，越详细越加分】

Answer 1

f(x)=[2sin^2(x+π/4)]-√3·cos2x
=2[sin(x+π/4)]^2- √3·cos2x
=2*(sinxcosπ/4+osxsinπ/4)^2-√3·cos2x
=(sinx+cosx)^2-√3·cos2x
=1+2sinxcosx-√3·cos2x
=1+sin2x-√3·cos2x
=1+2(sin2x*cosπ/3-cos2x*sinπ/3)
=1+2sin(2x-π/3
故最大值是3,最小值是-1

Answer 2

你能玩不，我才小学六年家

Answer 3

我就帮你化，答案就算了吧

2sin^2(x+π/4)-√3·cos2x
=2sin^2(x+π/4)-√3·sin(2x+π/2）
=2sin^2(x+π/4)-2√3·sin(x+π/4)sin(x+π/4)
=4sin(x+π/4)[1/2sin(x+π/4)-√3/2sin(x+π/4)]
=4sin(x+π/4)sin(x+π/4-π/3)
=4sin(x+π/4)sin(x-π/6)

Answer 4

∵cos2x=1-2sin^2x∴ 原式=1-cos2(x+@/4)-√3cos2x
=1-cos(2x+@/2)-√3cos2x
=1- (sin2x+√3cos2x)
﹉﹉﹉﹉﹉﹉﹉﹉
sin2x的倍数为1，√3cos2x倍数为√3 ，构造直角三角形，以1，√3为直角边算出斜边，将括号里边的两项提出斜边2就可以化成你要的形式
sin2x+√3cos2x=2(1/2sin2x+√3/2cos2x) 下面的总会了吧！够简单的1/2=cos60° √3/2=sin60°
2(1/2sin2x+√3/2cos2x)=2(sin2xcos60°+cos2xsin60°)=2sin(2x+60°)

核心:二倍角公式，两角和的正余弦公式的变换使用