Question

数学题求大佬帮忙

数学题求大佬帮忙麻烦过程详细一点 人比较蠢 谢谢

Answer 1

T=π

追问

谢谢啦(●'◡'●)ﾉ❤

追答

Answer 2

(1)
f(x)
= (sinx)^2 -[sin(x-π/6)]^2
=(1/2)(1-cos2x) - (1/2) [1 - cos(2x-π/3) ]
= (1/2)[ cos(2x-π/3) -cos2x]
= (1/2)[ (1/2)cos(2x)+(√3/2)sin2x -cos2x]
= (1/2)[ (√3/2)sin2x -(1/2)cos(2x)]
= (1/2)sin(2x-π/6)
最小正周期=π
(2)
x∈[-π/3 , π/4]
f(x) = (1/2)sin(2x-π/6)
max f(x)= f(π/4) = (1/2)sin(π/3) =√3/4
min f(x)= f(-π/6) = (1/2)sin(-π/2) =-1/2

追问

写的超级清楚诶！谢谢啦( 。ớ ₃ờ)ھ