已知函数f(x)=sin(π⼀2-x)-√3sin(π-x)求解

(1) f(x)=sin(π/2-x)-√3sin(π-x)
=2(1/2cosx-√3/2sinx)
=2(sinπ/6cosx-sinxcosπ/6)
=2sin(π/6-x)
T=2π/w=2π
f(x)max=2
f(x)min=-2
∴f(x)∈[-2,2]

2) f(a-π/3)=-2/3
2sin(π/6-a+π/3)=2sin(π/2-a)=2cosa
∴cosa=-1/3
∴cos2a/(1+cos2a-sin2a)=（2cos²a-1）/(1+2cos²a-1-2sinacosa)
sina=√(1-cos²a)=2√2/3
代入后原式=1/2-√2

f(x)=sin(π/2-x)-√3sin(π-x)

=cosx+√3sinx

=2sin(x+π/6)
∴最小正周期是T=2π，值域是[-2，2]
f(a-π/3)=2sin(a-π/3+π/6)=2sin(a-π/6)=-2/3
∴sin(a-π/6)=-1/3
∴cos(a-π/6)=-2√2/3
于是可求得sina与cosa的值。
∴cos2a/[1+cos2a-sin2a]
=cos2a/[1+2cos²a-1+2sinacosa]
=[cos²a-sin²a]/[2cosa(cosa+sina)]
=(cosa-sina)/(2cosa)可以求得其值