Question

已知向量a=(sin(π/2+x),cos(π-x)),向量b=(cosx,-sinx),函数f(x)=向量a*向量b. （1）求函数最小正周期

2，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

a=(cosx,-cosx),b=(cosx,-sinx)
(1) f(x)=ab=cos²x+sinxcosx
=(1+cos2x)/2 +(1/2)sin2x
=(√2/2)[(√2/2)sin2x +(√2/2)cos2x] +1/2
=(√2/2)sin(2x+π/4)+1/2
最小正周期为T=2π/2=π
(2) f(A)=(√2/2)sin(2A+π/4)+1/2=1
sin(2A+π/4)=√2/2
又A为锐角，从而 2A+π/4=3π/4，A=π/4
所以 由正弦定理，a/sinA=b/sinB，得
AC=b =asinB/sinA=√6

Answer 2

向量a=(sin(π/2+x)，cos(π-x))=(cosx，-cosx)
向量b=(cosx，-sinx)
f(x)=cos^2x+sinxcosx
=1/2cos2x+1/2sin2x+1/2
=√2/2sin(2x+π/4)+1/2
∴T=2π/2=π

f(A)=√2/2sin(2A+π/4)+1/2=1
sin(2A+π/4)=√2/2
0<A<π/2
π/4<2A+π/4<5π/4
∴2A+π/4=3π/4，
A=π/4
BC=a=2
AC=b=asinB/sinA
=(2×√3/2)/√2/2
=√6