求y=1+2sinx-2cos^2x的最大值和最小值
1个回答

y=1+2sinx-2cos^2x

=sin^2x+2sinx-cos^2x

=sin^2x+2sinx-(1-sin^2x)

=2sin^2x+2sinx-1

=(√2*sinx+√2/2)^2-3/2

当(√2*sinx+√2/2)^2=0时,y=-3/2最小;

当sinx=1,y=3最大;

相关问题