Question

y=sin^2+sin^2(x+π/3)+sin^2(X-π/3),并求出y的最大值

Answer 1

是 y=(sinx)^2+[sin(x+π/3)]^2+[sin(x-π/3)]^2

由倍角公式得

y=[1-cos(2x)]/2+[1-cos(2x+2π/3)]/2+[1-cos(2x-2π/3)]/2

=3/2-1/2*[cos(2x)+cos(2x+2π/3)+cos(2x-2π/3)]

=3/2-1/2*[cos(2x)+2*cos(2x)cos(2π/3)]

=3/2-1/2*[cos(2x)+2*cos(2x)*(-1/2)]

=3/2 ,

y 为定值,因此 y 的最大值为 3/2 .