已知:0<α<π/2<β<π,cos(β-π/4)=1/3,sin(α+β)=4/5 (1)求sin2β (2)求cos

已知:0<α<π/2<β<π,cos(β-π/4)=1/3,sin(α+β)=4/5 (1)求sin2β (2)求cos(α+π/4)的值
qq298008 1年前 已收到1个回答 举报

龙骨源 幼苗

共回答了18个问题采纳率:83.3% 举报

cos(β-π/4)=1/3
cos2(β-π/4)=2[cos(β-π/4)]^2-1=-7/9
cos(2β-π/2)=-7/9
cos(π/2-2β)=-7/9
sin2β=-7/9
cos(β-π/4)=1/3,sin(α+β)=4/5,0<α<π/2<β<π
sin(β-π/4)=2√2/3
cos(α+β)=-3/5
cos(α+π/4)=cos[(α+β)+(π/4-β)]
=cos(α+β)cos(π/4-β)-sin(α+β)sin(π/4-β)
=cos(α+β)cos(β-π/4)+sin(α+β)sin(β-π/4)
=(8√2-3)/15

1年前

1
可能相似的问题
Copyright © 2024 YULUCN.COM - 雨露学习互助 - 16 q. 0.024 s. - webmaster@yulucn.com