Question

已知函数f(x)=sin^2x+sinxcosx (1)求f( x)的最大值及取得最大值时对应的X的值 (2)求该函数的单调递增区间

Answer 1

f(x)=(sinx)^2+sinxcosx=(1-cos2x)/2+(sin2x)/2
=1-cos2x/2+sin2x/2
=1+(sin2x-cos2x)/2
=1+(sin2xcos45-cos2xsin45)/根号2
=1+sin(2x-45)/根号2
最大值是:1+1/根号2
对应的x值满足:sin(2x+45)=1
x=180k+22.5
递增区间:360k-90<=2x+45<=360k+90
180k-67.5<=x<=180k+22.5
自己算算仔细,将度数转成弧度比较好

Answer 2

f(x)=sin