sina=x,cosb=y,cos(a+b)=-3/5,求X的取值范围
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sin(a+b)=√(1-9/25)=4/5
sinacosb+cosasinb=4/5
cosacosb-sinasinb=-3/5
(tana+tanb)/(1-tanatanb)=-4/3
3tana+3tanb=-4+4tanatanb
tana(4tanb-1)=4+3tanb
tana=(4+3tanb)/(4tanb-1)
=(4cosb+3sinb)/4sinb-cosb)
(1+tan^2a)(4sinb-cosb)^2=16sin^2b-8sinbcosb+cos^2b+
16cos^2b+24sinbcosb+9sin^2b
=17+8sin2b+8sin^2b
=17+8sin2b+4-4cos2b
=21+4(2sin2b-cos2b)
sina=x=tana/√(1+tan^2a)
(4cosb+3sinb)(4sinb-cosb)=16sinbcosb+12sin^2b-4cos^2b-3sinbcosb
=13sinbcosb-4cos2b+4-4cos2b
=(13sin2b)/2-8cos2b+4
x=(6.5sin2b-8cos2b+4)/(21+4(2sin2b-cos2b))
=[4+√106.25sin(2b-A)]/[21+4√5sin(2b-B)]
A=arctan(8/6.5)
B=arctan(1/2)
很搞怪的题
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