Question

已知向量a=(sin(π/2+x),cos(π-x)),向量b=(cosx,-sinx),函数f(

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

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