摘要:11.已知sin(π-α)-cos(π+α)=.求下列各式的值: (1)sin α-cos α, (2)sin3+cos3. 解 由sin(π-α)-cos(π+α)=. 得sin α+cos α=.① 将①式两边平方.得1+2sin α·cos α=. 故2sin α·cos α=-. 又<α<π.∴sin α>0.cos α<0. ∴sin α-cos α>0. (1)(sin α-cos α)2=1-2sin α·cos α=1-=. ∴sin α-cos α=. (2)sin3+cos3=cos3α-sin3α =(cos α-sin α)(cos2α+cos α·sin α+sin2α) =×=-.

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