已知sin(a+3π/4)=5/13,cos(π/4-b)=4/5,且-π/4<a<π/4,π/4<b<3π/4求cos
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已知sin(a+3π/4)=5/13,cos(π/4-b)=4/5,且-π/4<a<π/4,π/4<b<3π/4求cos(a-b)的值
由-π/4<a<π/4,π/4<b<3π/4,得:
π/2<a+3π/4<π,-π/2<π/4-b<0,
由sin(a+3π/4)=5/13,cos(π/4-b)=4/5,得:
cos(a+3π/4)=-12/13,sin(π/4-b)=-3/5,
所以
cos(a-b)=cos[π+(a-b)]=cos[(a+3π/4)+(π/4-b)]
=cos(a+3π/4)*cos(π/4-b)-sin(a+3π/4)*sin(π/4-b)
=-12/13*4/5+5/13*3/5=-33/65.
π/2<a+3π/4<π,-π/2<π/4-b<0,
由sin(a+3π/4)=5/13,cos(π/4-b)=4/5,得:
cos(a+3π/4)=-12/13,sin(π/4-b)=-3/5,
所以
cos(a-b)=cos[π+(a-b)]=cos[(a+3π/4)+(π/4-b)]
=cos(a+3π/4)*cos(π/4-b)-sin(a+3π/4)*sin(π/4-b)
=-12/13*4/5+5/13*3/5=-33/65.
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