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已知函数f (x)=1/2sinx+根号3/2cosx,x属于R.
(1)求函数f(x)的最小正周期和值域
(2)记△ABC的内角A,B,C 的对边分别是a,b,c,若f(A)=根号3/2,a=(根号3)b/2,求角c的值
人气:301 ℃ 时间:2019-10-19 12:58:16
解答
(1)
f(x)=(1/2)sinx+(√3/2)cosx
=sin(x+π/3)
最小正周期=2π
值域=[-1,1]
(2)
f(A)=√3/2
sin(A+π/3)=√3/2
A+π/3=2π/3
A=π/3
a=√3( b/2)
a/sinA = b/sinB
a/sin(π/3) = (2a/√3)/sinB
sinB =1
B=π/2
C= π-A-B
= π/6
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