设函数f(x0=四年(x+π/6)+2sin^2x/2.记△ABC的内角A,B,C的对边a,b,c,若f(A)=1,a=1,c=根号3,求b 值
问题描述:
设函数f(x0=四年(x+π/6)+2sin^2x/2.记△ABC的内角A,B,C的对边a,b,c,若f(A)=1,a=1,c=根号3,求b 值
答
f(x)=sin(x+π/6)+1-cosx=(根号3/2)sinx+(1/2)cosx+1-cosx=(根号3/2)sinx-(1/2)cosx+1=sin(x-π/6)+1又f(A)=1 A∈(0,π)所以sin(A-π/6)=0 A=π/6又c=3 a=1由余弦定理知1=b*2+c*2-2bc*cos(π/6) b*2-3b+...