若sin(π/4-α)=-4/5 , sin(3π/4+β)=5/13 , 0<α<3π/4 , -π/4<β<π/4 , 求cos(α+β)的值.

来源：百度知道 编辑：UC知道 时间：2024/05/10 12:06:19
20分

sin(3π/4+β-π/4+α)=sin(π/2+α+β)=-cos(α+β)

cos(3π/4+β)=-12/13 cos(π/4-α)=3/5 [

sin(3π/4+β-π/4+α)=cos(π/4-α)sin(3π/4+β)-sin(π/4-α)cos(3π/4+β)=(5/13)*(3/5)-(-4/5)*(-12/13)=-33/65

整体带入法简单哦

-33/65
解:sin(3π/4+β-π/4+α)=sin(π/2+α+β)=cos(α+β)
因为sin(π/4-α)=-4/5 , sin(3π/4+β)=5/13 , 0<α<3π/4 , -π/4<β<π/4
所以cos(π/4-α)=3/5,cos(3π/4+β)=-12/13
则cos(α+β)=sin(3π/4+β-π/4+α)
=sin(3π/4+β)*cos(π/4-α)-cos(3π/4+β)*sin(π/4-α)
=-33/65