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sinα3.在△ABC中,若sinAsinB<cosAcosB,试判断△ABC的形状
问题描述:
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
答
1.======-1
2.======(根号10-2根号2)/6
3.======0<cos(A+B)钝角三角形
自己看着办吧,不知对不对,手算的
没检验·······
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...