①已知cos(π/4-a)=3/5,sin(3π/4+b)=5/13,其中π/4

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①已知cos(π/4-a)=3/5,sin(3π/4+b)=5/13,其中π/4
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①已知cos(π/4-a)=3/5,sin(3π/4+b)=5/13,其中π/4
①已知cos(π/4-a)=3/5,sin(3π/4+b)=5/13,其中π/4

①已知cos(π/4-a)=3/5,sin(3π/4+b)=5/13,其中π/4
1.因为(3π/4+b)-(π/4-a)=b+a+π/2
所以cos[(a+b)+π/2]=cos(a+b)*cosπ/2-sin(a+b)*sinπ/2=-sin(a+b)
所以sin(a+b)=-cos[(a+b)+π/2]=-cos[(3π/4+b)-(π/4-a)]
=-[cos(3π/4+b)*cos(π/4-a)+sin(3π/4+b)*sin(π/4-a)]
由π/4

  1. cos(π/4-a)=3/5,
    可以求出sin(π/4-a)=±4/5
    又π/4<a<3π/4, -π/2<π/4-a<0
    ∴sin(π/4-a)=-4/5
    同理:
    sin(3π/4+b)=5/13;
    cos(3π/4+b)=-12/13

    注意:a+b=[(3π/4+b)-(π/4-a)]-...

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    1. cos(π/4-a)=3/5,
      可以求出sin(π/4-a)=±4/5
      又π/4<a<3π/4, -π/2<π/4-a<0
      ∴sin(π/4-a)=-4/5
      同理:
      sin(3π/4+b)=5/13;
      cos(3π/4+b)=-12/13

      注意:a+b=[(3π/4+b)-(π/4-a)]-π/2
      ∴sin(a+b)=sin{[(3π/4+b)-(π/4-a)]-π/2}=-cos{[(3π/4+b)-(π/4-a)]=-cos(3π/4+b)cos(π/4-a)]-sin(3π/4+b)sin(π/4-a)]
      =-(-12/13)*(3/5)-(5/13)(-4/5)=56/65

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