有两道高中数学题不会做，求过程

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(2)tanα=√3→宽渣(secα)^2=1+(tanα)^2=4，α∈第3象限，sedα=-2
sinα-cosα=cosα(tanα-1)=(1/慎槐悄secα)(tanα-1)=-(1/2)(√3-1)=(1-√3)/明宴2
故应选D
(3)(sinα-cosα)^2=1/4→sinαcosα=3/8
(sinα)^3-(cosα)^3=(sinα-cosα)[(sinα)^2+sinαcosα+(cosα)^2]=(-1/2)(1+3/8)=-11/16

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（2） sin(a) = -根稿枯姿败兄号3 /2 cos(a) = -1/2
sin(a) - cos(a) = (1-根号3 )/2
D
(3) (sina - cosa)^2 = 1-2*sina*cosa =1/4
sina*cosa = 3/8
sina^3 - cosa^3 = (sina - cosa)(sina^2 + sina*cosa + cosa^2) =-1/2 * (1+3/8) = -11/键绝8

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