已知派/4

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已知派/4已知派/4已知派/4π/4先求sin(π/4-a)=-3/5cos(3π/4+b)=-12/13然后cos(3π/4+b)cos(π/4-a)+sin(3π/4+b)sin(π/4-a)=c

已知派/4
已知派/4

已知派/4
π/4

先求sin(π/4-a)=-3/5 cos(3π/4+b)=-12/13 然后cos(3π/4+b)cos(π/4-a)+sin(3π/4+b)sin(π/4-a)=cos(π/4+a+b)=-sin(a+b)=-63/65 所以sin(a+b)=63/65

-π/2<π/4-a<0,sin(π/4-a)=-3/5<0;3π/4<3π/4+b<π,cos(3π/4+b)=-12/13<0;
sin(a+b)=-cos(π/2+a+b)=-cos(3π/4+b)cos(π/4-a)-sin(3π/4+b)sin(π/4-a)
=12/13×4/5+3/5×5/13
=63/65

由-pi/12sin(pi/4-a)<0得sin(pi/4-a)=-3/5
同理cos(3pi/4+b)=-12/13
所以
sin(a+b)=-cos[(3pi/4+b)-(pi/4-a)]=-cos(3pi/4+b)*cos(pi/4-a) - sin(3pi/4+b)*sin(pi/4-a)=63/65.

sin(α+β)
=sin(π+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)
=5/13×4/5-(-3/5)×(-12/13)
=20/65-36/65
=-16/65