已知sin(a-π/6)=3/5,a∈(0,π/2),求sina
答案:2 悬赏:80
解决时间 2021-03-17 19:51
- 提问者网友:刀枪不入
- 2021-03-16 19:04
已知sin(a-π/6)=3/5,a∈(0,π/2),求sina
最佳答案
- 二级知识专家网友:一场云烟
- 2021-03-16 20:15
a∈(0,π/2)
a-π/6∈(-π/6,π/3)
sin(a-π/6)=3/5,所以 cos(a-π/6)=4/5
sina=sin[(a-π/6)+π/6]
=sin(a-π/6)sin(π/6)+cos(a-π/6)cos(π/6)
=(3/5)*(1/2)+(4/5)*(√3/2)
=(3+4√3)/10
a-π/6∈(-π/6,π/3)
sin(a-π/6)=3/5,所以 cos(a-π/6)=4/5
sina=sin[(a-π/6)+π/6]
=sin(a-π/6)sin(π/6)+cos(a-π/6)cos(π/6)
=(3/5)*(1/2)+(4/5)*(√3/2)
=(3+4√3)/10
全部回答
- 1楼网友:绝望伪装
- 2021-03-16 21:06
cos(a-π/6)+sina=cosacosπ/6+sinasinπ/6+sina
=(√3/2)cosa+(3/2)sina=4√3/5
∴(1/2)cosa+(√3/2)sina=4/5
sin(x+7π/6)=sinacos7π/6+sin7π/6cosa
=(-√3/2)sina+(-1/2)cosa=-4/5
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