已知a,b∈(3π/4,π),sin(a+b)= - 3/5,sin(b-π/4)=12/13,求cos(a+π/4)

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已知a,b∈(3π/4,π),sin(a+b)= - 3/5,sin(b-π/4)=12/13,求cos(a+π/4)

已知a,b∈(3π/4,π),sin(a+b)= - 3/5,sin(b-π/4)=12/13,求cos(a+π/4)
已知a,b∈(3π/4,π),sin(a+b)= - 3/5,sin(b-π/4)=12/13,求cos(a+π/4)

已知a,b∈(3π/4,π),sin(a+b)= - 3/5,sin(b-π/4)=12/13,求cos(a+π/4)
因为a,b∈(3π/4,π)
所以a+b∈(3π/2,2π)
b-π/4∈(π/2,3π/4)
所以cos(a+b)=4/5
cos(b-π/4)=-5/13
cos(a+π/4)=cos[(a+b)-(b-π/4)]
=cos(a+b)cos(b-π/4)+sin(a+b)sin(b-π/4)
=4/5*(-5/13)+(-3/5)(12/13)
=-56/65

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http://zhidao.baidu.com/question/84723797.html?fr=qrl&cid=983&index=1

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