1.已知tana=1/3,tanb=-1/4求tan(2a-2b).2.已知sin(a+π/6)=1/3求sin(π/6

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1.已知tana=1/3,tanb=-1/4求tan(2a-2b).2.已知sin(a+π/6)=1/3求sin(π/6-2a)

1.由 tan(α－β)＝(tanα－tanβ)/(1＋tanα•tanβ)和tan2A=2tanA/(1－tan2A),得
tan(2a-2b)=(tan2a-tan2b)/(1+tan2a tan2b)
=[2tana/(1-tan²a)-2tanb/(1-tan²b)]/{1+4tana tanb/[(1-tan²a)(1-tan²b)]} ①
将①式分子分母同乘以(1-tan²a)(1-tan²b),化简,得
tan(2a-2b)=[2tana(1-tan²b)-2tanb(1-tan²a)]/[ (1-tan²a)(1-tan²b)-4tana tanb]
=[2tana-2tanb+2tana tanb(tana-tanb)]/(tan²a tan²b-tan²a-tan²b+4tana tanb+1)
=2(1+tana tanb)(tana-tanb)/[(tana tanb+1)²-(tana-tanb)²] ②
将已知值代入②,计算,得
tan(2a-2b)=77/36
2.由sin(π-A)＝sinA和sin(π/2+A)＝cosA,得
sin(π/6-2a)=sin[π-(π/6-2a)]=sin[(2a+π/3)+ π/2]=cos(2a+π/3)
再由cos2B＝1－2sin2B,得
sin(π/6-2a)= cos(2a+π/3)=1-2sin²(a+π/6)=1-2*(1/3)²=7/9

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