已知函数f(x)=sin(2x+π/6)-cos(2x+π/3)+2cos²x. (1)求f(π/12)的值;
答案:1 悬赏:80 手机版
解决时间 2021-02-17 04:54
- 提问者网友:树红树绿
- 2021-02-16 11:54
已知函数f(x)=sin(2x+π/6)-cos(2x+π/3)+2cos²x. (1)求f(π/12)的值;
最佳答案
- 五星知识达人网友:不甚了了
- 2021-02-16 12:53
f(x)=sin(2x+π/6)-cos(2x+π/3)+2cos²x
= sin2x cosπ/6+cos2x sinπ/6-[ cos2x cosπ/3- sin2x sinπ/3] +2cos²x
=√3 sin2x+2cos²x
=√3 sin2x+1+cos2x
=2 sin(2x+π/6)+1,
f(π/12)=2 sin(2*π/12+π/6)+1=√3+1.
f(x)的最大值是3,此时2x+π/6=2kπ+π/2,
x=kπ+π/6,k∈Z.
= sin2x cosπ/6+cos2x sinπ/6-[ cos2x cosπ/3- sin2x sinπ/3] +2cos²x
=√3 sin2x+2cos²x
=√3 sin2x+1+cos2x
=2 sin(2x+π/6)+1,
f(π/12)=2 sin(2*π/12+π/6)+1=√3+1.
f(x)的最大值是3,此时2x+π/6=2kπ+π/2,
x=kπ+π/6,k∈Z.
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