Question

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

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

答案：16/65

要过程。十分感谢。

Answer 1

解: ∵ π/4＜α＜3π/4
∴ -π/2<π/4-α<0
sin(π/4-α)=-4/5 ,cos(π/4-α)=3/5
∵ 0＜β＜π/4
∴ 3π/4 <3π/4+β<π
sin(3π/4+β)=5/13 ,cos(3π/4+β)=-12/13
cos(α-β)=-cos[π-(α-β)] = - cos[(π/4-α)+(3π/4+β)] = - [cos(π/4-α)cos(3π/4+β)-sin(π/4-α)sin(3π/4+β)] = - [3/5×(-12/13)-(-4/5)×5/13] = 16/65.

Answer 2

π/4<α<3π/4
则-3π/4<-α<-π/4
-3π/4+π/4<π/4-α<-π/4+π/4
-π/2<π/4-α<0
所以cos(π/4-α)=3/5
同理
0<β<π/4
0+3π/4<β+3π/4<π/4+3π/4
3π/4<β+3π/4<π
cos(β+3π/4)=-12/13
而cos(α-β)=cosαcosβ+sinαsinβ
代入得16/65
记得采纳哦

Answer 3

π/4＜α＜3π/4,
-π/2＜π/4-α＜0
sin(π/4-α)=-4/5
cos(π/4-α)=3/5
0＜β＜π/4
3π/4＜3π/4+β＜π
sin(3π/4+β)=5/13
cos(3π/4+β)=-12/13
cos(α-β)
=-cos[π-(α-β)]
=-cos[(π/4-α)+(3π/4+β)]
=-[cos[(π/4-α)*cos(3π/4+β)-sin(π/4-α)*sin(3π/4+β)]
=-[3/5*(-12/13)-(-4/5)*5/13]
=-[-16/65]
=16/65