1楼:
sin(α+π/6)=cos(π/2-(a+π/6))=cos(π/3-a)=1/3
如令π/3-a=x,则2π/3-2α=2x
cos2x=2cos2x-1=2*1/9-1=-7/9
2楼:匿名用户
cos(2π/3-2α)
=-cos[π-(2π/3-2α)]
=-cos(π/3+2α)
=-cos[2(π/6+α)]
=- =-(1-2/9)
=-7/9
若sin(π/6-α)=1/3,则cos(2π/3+2α)的值? 请针对每一步做说明,清楚一点。
3楼:匿名用户
解:ducos(2π
zhi/3+2α
)= - cos(πdao-(2π/3+2α回))=- cos(π/3-2α)
=- cos(2α-π/3) (公式答cos2α=1-2(sinα)^2)
=- (1-2(sin(α-π/6))^2)因为 sin(π/6-α)=1/3 所以(sin(α-π/6))^2=1/9
故cos(2π/3+2α)= -(1-2*1/9)=-7/9
4楼:一仇冒斩
sin(π/6-α)=1/3,
cos(π/3-2α)=1-2sin(π/6-α)sin(π/6-α)=1-2/9=7/9
cos(2π/3+2α)=-cos(π-π/3+2α)=-7/9
5楼:逆天者天逆之
π=180°,然后你可以用计算器算
已知sin(π/6+α)=1/3,则cos(2π/3-2α)=?
6楼:雨叶泪水
^cos[π袭/2-(π
bai/6+α)
du]=sin(π/6+α)zhi=1/3所以cos(πdao/3-α)=1/3 利用倍角公式则cos2(π/3-α)=cos(2π/3-2α)=2[cos(π/3-α)]^2-1=2*(1/9)-1=-7/9
7楼:匿名用户
sin(π/6+α)=cos(π/2-π/6-α)=cos(π/3-α)=1/3
cos(2π/3-2α)=cos2(π/3-α)=2cos^2(π/3-α)-1=2/9-1=-7/9
8楼:蘅芜清芬
cos(2π/3-2α)=cos(2α-2π/3)=cos(2a-π+π/3)=-cos(2a+π/3)=-(1-2sin(2a+π/6)^2)=-7/9
若sin(π/6-α)=1/3,则cos(2π/3+2α)的值
9楼:我不是他舅
cos(2π/3+2α)
=-cos[π-(2π/3+2α)]
=-cos(π/3-2α)
=-cos[2(π/6-α)]
=- =-(1-2/9)
=-7/9
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