在三角形ABC中,角A,B,C所对的边分别为a,b,c,a=2倍根号3,b=2,cosA=-1/2(1)求角B的大小(2)若f(x)=cos2x+bsin的2次方(x+B)求函数的最小正周期和单调递增区间
问题描述：
在三角形ABC中,角A,B,C所对的边分别为a,b,c,a=2倍根号3,b=2,cosA=-1/2
(1)求角B的大小
(2)若f(x)=cos2x+bsin的2次方(x+B)求函数的最小正周期和单调递增区间
答
(1)
cosA=-1/2 A是纯角
sinA=√3/2
a/sinA=b/sinB
sinB=b*sinA/a=2*√3/2 /2√3=1/2
B=π/6
(2)
f(x)=cos2x+b(sin(x+π/6))^2
=cos2x+2*(1-cos2(x+π/6))/2
=cos2x +1-cos(2x+π/3)
=cos2x+1-(cos2xcosπ/3-sin2xsinπ/3)
=cos2x+1-cos2x /2 +√3sin2x/2
=√3sin2x/2+cos2x /2 +1
=sin2xcosπ/6+cos2xsinπ/6+1
=sin(2x+π/6)+1
所以最小正周期T=2π/2=π
当 2kπ-π/2