已知sin(5pi+a)=-3/5,且a属于(pi/2,pi)又tan(b+pi/4)=3.
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sin(5pi+a)=-3/5
-sina=-3/5
sina=3/5
∵a属于(pi/2,pi)
∴cosa=-√(1-sin^2a)=-4/5
tana=sina/cosa=-3/4
tan(b+pi/4)=3
(tanb+tanpi/4)/(1-tanbtanpi/4)=3
(tanb+1)/(1-tanb)=3
tanb+1=3-3tanb
tanb=1/2
tan(a-b)
=(tana-tanb)/(1+tanatanb)
=(-3/4-1/2)/(1+(-3/4)*1/2))
=-5/4 / 5/8
=-2
sin2a=2sinacosa=2*3/5*(-4/5)=-24/25
cos2a=2cos^2a-1=2*(-4/5)^2-1=7/25
sin(2a+pi／3)
=sin2acospi／3+cos2asinpi／3
=-24/25*1/2 +7/25*√3/2
=(7√3-24)/50
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