Question

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

Answer 1

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

Answer 2

f(x)=sin(π/2+x)*cosx-sinx*cos(π-x)
=cos^2x+sinx*cosx)
=1/2*(1+cos2x)+1/2*sin2x
=√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
因A为锐角，所以2A+π/4=3π/4
A=π/4
由正弦定理BC/sinA=AC/sinB
2/sinπ/4=AC/sinπ/3
解得AC=√6

Answer 3

解：f(A)=sin(π/2+A)*cosA-sinx*cos(π-A)
=cosA*cosA+sinA*cosA=1
所以： sinA*cosA=1-cosA*cosA= sinA*sinA
即：sinA*cosA-sinA*sinA=sinA*（cosA-sinA）=0
因为： A为锐角，
所以：sinA#0，cosA=sinA
即 A=π/4
AC/sinB=BC/sinA
所以AC=2sin(π/3)/(π/4)=根号6/2

Answer 4

f(x)=sin(π/2+x)*cosx-sinx*cos(π-x)=(cosx)^2+sinxcocx=1/2(cos2x+sin2x)+1/2,因为f(A)=1，所以cos2A+sin2A=1,A=45度，AC=√6