Question

求详细过程

Answer 1

（1） f(x)=2√3cosxsinx-cos²x+sin²x
=√3sin2x-cos2x
=2(sin2xcosπ/6-cos2xsinπ/6)
=2sin(2x-π/6)
增区间满足：
2kπ-π/2≤2x-π/6≤2kπ+π/2
2kπ-π/3≤2x≤2kπ+2π/3
kπ-π/6≤x≤kπ+π/3
即增区间为【kπ-π/6，kπ+π/3】，k∈Z
(2)
f(x)=2sin(2x-π/6)
f(a/2)=2sin(a-π/6)=1/2
sin(a-π/6)=1/4
因为π/6<a<2π/3
所以
0<a-π/6<2π/3-π/6=π/2
从而
cos(a-π/6)=√（4²-1）/4=√15/4
所以
sina
=sin(a-π/6+π/6)
=sin(a-π/6)cosπ/6+cos(a-π/6)sinπ/6
=1/4 ×√3/2+√15/4×1/2
=(√3+√15)/8

追问

辛苦了

Answer 2

我看看

追问

仔细看看

追答

高一数学吧

Answer 3

追问

化简的过程

追答

采纳采纳

追问

？？？？

追答

。。。。

错了啊同学，楼上化简有问题

Answer 4