设函数F（x）=cos(2x-π/3)-2sin^2x
答案:1 悬赏:30 手机版
解决时间 2021-01-20 07:24
- 提问者网友：轻浮
- 2021-01-19 12:17
设函数F（x）=cos(2x-π/3)-2sin^2x
最佳答案
- 五星知识达人网友：鱼忧
- 2021-01-19 13:38
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
希望能帮到你O(∩_∩)O追问2cos(2x-π/6)*cos(π/6)-1什么意思可以用中文说下么？追答你是说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]
=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
希望能帮到你O(∩_∩)O追问2cos(2x-π/6)*cos(π/6)-1什么意思可以用中文说下么？追答你是说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]
我要举报
如以上问答信息为低俗、色情、不良、暴力、侵权、涉及违法等信息，可以点下面链接进行举报！
大家都在看
推荐资讯