Question

已知cos(π/4-β)=3/5,sin(3π/4+α)=5/13，求cos(α-β)

且-π/4<α<π/4，π/4<β<3π/4.

Answer 1

因为-π/4<α<π/4

所以π/2<3π/4+α<π

所以cos(3π/4+α)=-12/13

同理，因为π/4<β<3π/4

所以-3π/4<-β<-π/4

所以-π/2<π/4-β<0

所以sin(π/4-β)=-4/5

cos[(π/4-β)+(3π/4+α)]=cos(π/4-β)cos(3π/4+α)-sin(π/4-β)sin(3π/4+α)=-16/65=cos(α-β+π)

cos(α-β+π)=-cos(α-β)=16/65

Answer 2

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