设函数F（x）=cos(2x-π/3)-2sin^2x

来源：学生作业帮编辑：神马作文网作业帮分类：数学作业时间：2024/11/12 02:19:36

设函数F（x）=cos(2x-π/3)-2sin^2x
求最小正周期和单调递增区间
（2）三角形ABC,角A,B,C所对边分别是a,b ,c.且F(B)=1/2,B=1,C=根号3.求a的值.

F（x）=cos(2x-π/3)-2sin^2x=cos(2x-π/3)+cos2x-1
=2cos(2x-π/6)*cos(π/6)-1
=√3cos(2x-π/6)-1
(1) 则最小正周期T=2π/2=π
单增区间2x-π/6∈[2kπ-π/2,2kπ+π/2]
x∈[kπ-π/6,kπ+π/3]
(2) F(B)=√3cos(2B-π/6)-1=1/2 cos(2B-π/6)=√3/2
2B-π/6=π/6 B=π/6
由余弦定理b²=a²+c²-2accosB
则1=a²+3-2a*√3*√3/2
a²-3a+2=0 (a-1)(a-2)=0
解得a=1或2
再问： 2cos(2x-π/6)*cos(π/6)-1什么意思可以用中文说下么？
再答：你是说cos(2x-π/3)+cos2x-1=2cos(2x-π/6)*cos(π/6)-1 用到和差化积公式: cosa+cosb=2cos[(a+b)/2]cos[(a-b)/2]

已知函数f(x)=cos(2x-π/3)+sin^2x-cos^2x,设函数g(x)=[f(x)]^2+f(x),求g( 设函数f(x)=cos(2x+π/3)+sin^2x-1/2 设函数f(x)=cos(2x+π/3)+sin^2 X 设函数f(x)=cos(2x+π/3)+sin^2x 急.设函数f(x)=cos(2x+π/3)+sin^2 X 设函数f(x)=cos(2x+π/3)+sin方x 1.设函数f(x)=cos(2x+π/3)+sin(平方)x 设函数f(x)=cos(2x+π/3)+sin²x 设函数f(x)=cos(2x+π/3)+sin²X 设函数f(x)=2+sin(3x+π/12)cos(x+π/6)+cos(3x+π/12)sin(x+π/6) 设函数f(x)=cos(2x+pai/3)+sin^2x 设函数f(x)=sin²x+2sin2x+3cos²x 化简