Question

已知函数FX=SIN(X+3/π)COS(X-π/3)-√3/4

已知函数FX=SIN(X+3/π)COS(X-π/3)-√3/4
1求FX的最小正周期
2求FX在区间[-π/6，π/2]上的最大值和最小值

要过程 感激不尽

Answer 1

1.f(x)=sin(π+x)cos(π-x)tan(π-x)/tan(π+x)cos(3π/2+x)
=(-sinx)(-cosx)(-tanx)/(tanx)(sinx)
=-cosx
2.若x是第二象限角，且cos（x-π/3）=1/5>0,sin（x-π/3）=2√6/5
cosx=cos[(x-π/3)+π/3]=cos(x-π/3)cosπ/3-sin(x-π/3)sinπ/3=1/5*1/2-2√6/5*√3/2=(1-6√2)/10
f(x)=-cosx=(6√2-1)/10

Answer 2

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