y=2sinx(sinx+cosx0最小正周期和最小值
1个回答

是y=2sinx(sinx+cosx)吧

记y=f(x)=2sinx(sinx+cosx)

f(x)=2(sinx)^2+2sinxcosx

=1-cos2x+sin2x

=2^0.5(sin2x/2^0.5-cos2x/2^0.5)+1

=2^0.5(sin2xcos(pi/4)-cos2xsin(pi/4))+1

=2^0.5sin(2x-pi/4)+1

其中2^0.5表示根号2

故f的最小正周期为pi

当x=k*pi-pi/8时,f取最小值1-2^0.5