已知函数f(x)=sin(π/2+x)*cosx-sinx*cos(π-x)在三角形ABC中,已知A为锐角,f(A)=1,

问题描述:

已知函数f(x)=sin(π/2+x)*cosx-sinx*cos(π-x)在三角形ABC中,已知A为锐角,f(A)=1,
BC=2,B=π/3,求AC边的长

f(x)=cos²x+sinxcosx
=1/2*(sin2x+cos2x)+1/2
=√2/2*sin(2x+π/4)+1/2
f(A)=√2/2*sin(2A+π/4)+1/2=1
sin(2A+π/4)=√2/2
2A+π/4=3π/4
A=π/4
a/sinA=b/sinB
2/(√2/2)=b/(√3/2)
所以AC=b=√6