Question

已知f(x)=cos3x/2cosx/2-sin3x/2sinx/2-2sinxcosx…

Answer 1

(1) f(x)=cos3x/2 cosx/2 -sin3x/2 sinx/2-2sinxcosx

=cos（3x/2 +x/2）-sin2x

=cos2x-sin2x

=-√2sin（2x-π/4）.

f(x)的最少正周期是 2π/2,即π.

(2) 当x∈[π/2,π]时,2x-π/4∈[3π/4,7π/4],

在[3π/4,7π/4]内,使f(x)=0的值是π,

由2x-π/4=π,得x=5π/8.

即函数f(x)的零点是x=5π/8.