问题标题：
已知cos(π/6+α)=1/2,则sin(2π/3+α)=?已知cos(π/2+α)=2/3,且α∈(-π/2,0),那么tan(3π/2+α)=?已知(π/3+α)=-1/3,则cos(7π/6-α)=?
更新时间：2024-04-28 08:00:49
问题描述：
已知cos(π/6+α)=1/2,则sin(2π/3+α)=?已知cos(π/2+α)=2/3,且α∈(-π/2,0),那么tan(3π/2+α)=?
已知(π/3+α)=-1/3,则cos(7π/6-α)=?
陶海军回答：
1.cos(π/6+α)=-sin(π/6+α+π/2)=-sin(2π/3+α)=1/2,所以sin(2π/3+α)=-1/2,2.因为α∈(-π/2,0)所以π/2+α∈(0,π/2)cos(π/2+α)=2/3sin(α)=-2/3cos(α)=√1-sin²α=√5/3tan(3π/2+α)=-cot(α)=-cos...
梁玉红回答：
第三题是sin(π/3+α)=-1/3，则cos(7π/6-α)=?抱歉打少了..sin(π/3+α)=-1/3=-cos[3π/2-(π/3+α)]=-1/3所以cos(7π/6-α)=1/3对么..？
陶海军回答：
很好！
