1樓：匿名使用者
cos[(π/4-a)-(3π/4+β)]
=cos(-π/2-a-β)
=cos(π/2+a+β)
=sin(a+β)
sin(a+β)=(3/5)(12/13)+(5/13)(4/5)
=56/65
2.cos(π/4-α)=3/5
π/4<α<3π/4 -3π/4<-α<-π/4
則-π/2<π/4-α<0
sin(π/4-α)<0
sin(π/4-α)=-√(1-(cos(π/4-α))^2)=-4/5
sin(3π/4+β)=5/13
0<β<π/4 3π/4<3π/4+β<π
cos(3π/4+β)<0
cos(3π/4+β)=-√(1-(sin(3π/4+β))^2)=-12/13
sin(α+β)=-cos(α+β+π/2)=-cos((3π/4+β)-(π/4-α))
=-[cos(3π/4+β)cos(π/4-α)+sin(3π/4+β)sin(π/4-α)]
=-[(-12/13)*3/5+5/13*(-4/5)]
=56/65
2樓：有誰不知道呢
cos〔3π/4+β-(π/4-α)〕=cos(π/2+α+β)=-sin(α+β)
=cos(3π/4+β)cos(π/4-α)+sin(3π/4+β)sin(π/4-α)
0<β<π/4<α<3π/4
所以3π/4<β+3π/4<2π，-π/2<α<0sin(3π/4+β)=5/13＞0所以3π/4<β+3π/4<π,cos(3π/4+β)＜0 sin(π/4-α)＜0=-12/13 =-4/5cos(3π/4+β)cos(π/4-α)+sin(3π/4+β)sin(π/4-α)
=-36/65-20/65=-56/65=-sin(α+β)所以sin(α+β)=56/65
已知0<β<π/4<α<3π/4,且cos(π/4-α)=4/5,sin(3π/4+β)=5/13
3樓：匿名使用者
見圖：若幫到你，請給好評哦！祝你學習進步！
已知0<α<π/4<β<3π/4,cos(π/4 -β)=3/5,sin(3π/4+α)=5/13,求sin(α+β)的值
4樓：匿名使用者
解：∵0<α<π/4<β<3π/4
==>-π/2<π/4-β<0,3π/4<3π/4+α<π
∴sin(π/4-β)<0,cos(3π/4+α)<0
∵cos(π/4-β)=3/5,sin(3π/4+α)=5/13
∴sin(π/4-β)=-√[1-(cos(π/4-β))^2]=-4/5
cos(3π/4+α)=-√[1-(sin(3π/4+α))^2]=-12/13
故sin(α+β)=sin[(3π/4+α)-(π/4-β)-π/2]
=-sin[(3π/4+α)-(π/4-β)]
=cos(3π/4+α)sin(π/4-β)-sin(3π/4+α)cos(π/4-β) (應用差角公式)
=(-12/13)(-4/5)-(5/13)(3/5)
=33/65。
已知π/4<α<3π/4，0<β<π/4，cos(π/4+α)=-3/5，sin(3π/4+β)=5/13，求sin(α+β)的值
5樓：匿名使用者
由π/4<α<3π/4，0<β<π/4，cos(π/4+α)=-3/5，sin(3π/4+β)=5/13，得出sin(π/4+α)=4/5，cos(3π/4+β)=-12/13。sin(α+β)=sin(π/4+α+3π/4+β-π)=-sin(π/4+α+3π/4+β)再化開得
sin(α+β)=-（4/5）*（-12/13）-（5/13）*（-3/5）=63/65
而且由π/4<α<3π/4，0<β<π/4，知道 π/4<α+β<π，sin(α+β)又怎麼可能是負數，很明顯答案錯了。
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