同角基本关系专题（4） θ α π
1个回答

7.已知tanα是方程x2+(2/cosα)x+1=0的根,求sin(α-π).

tan2α+(2/cosα)tanα+1=0

将tanα化成sinα/cosα,sin2α/cos2α + 2sinα/cos2α + 1 = 0

两边同乘cos2α,sin2α + 2sinα + cos2α ＝0

1+2sinα=0

sinα=-1/2

故sin(α-π)=-sinα=1/2

8.若cos(π/4-α)=m,求sin(π/4+α)、cos(3π/4+α).

sin(π/4+α) = -sin(-π/4-α) = -cos(π/4-α) = -m

cos(3π/4+α) = cos(π-(π/4-α)) = -cos(π/4-α) = -m 或者这样算：cos(3π/4+α) = sina(π/4+α+π/2)= sin(π/4+α) = -m

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