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1.已知cosα=-12/13,α∈(π/2,π),求tan(α-π/4)
2.已知sin(45°-α)=-2/3,π/4<α<π/2,求sinα
3.在△ABC中,若sinAsinB<cosAcosB,试判断△ABC的形状
人气:146 ℃ 时间:2020-07-10 21:40:07
解答
1.由cosα=-12/13,α∈(π/2,π)得 sinα=5/13,tanα=-5/12tan(α-π/4)=(tanα-tanπ/4)/(1+tanαtanπ/4)=(-5/12-1)/(1+(-5/12)) =-17/72..根据sin(45°-α)=-2/3 求出cos(45°-α)= √5 /3cos(2α)=sin(90...
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