若tan(π-α)=2,则2sin(3π+α)cos(5/2π+α)+sin(3/2π-α)sin(π-α)的值为?

来源：学生作业帮助网编辑：作业帮时间：2024/05/13 03:11:06

若tan(π-α)=2,则2sin(3π+α)cos(5/2π+α)+sin(3/2π-α)sin(π-α)的值为?
若tan(π-α)=2,则2sin(3π+α)cos(5/2π+α)+sin(3/2π-α)sin(π-α)的值为?

若tan(π-α)=2,则2sin(3π+α)cos(5/2π+α)+sin(3/2π-α)sin(π-α)的值为?
tan(π-α)=2 =>tanα=-2
2sin(3π+α)cos(5/2π+α)+sin(3/2π-α)sin(π-α)
=(-2sinα)(-sina)+(-cosa)sina
=2sin² α-cosasina
=2tan²acos²a-cos²atana
=cos²a[2x4+2]
=10cos²a
=10/sec²a
=10/[1+tan²a]
=10/5
=2

tanα=-2

原式=2(sinα)^2-cosαsinα=(2×sin²α-cosα×sinα)/(sin²α+cos²α)=(2tan²α-tanα)/(tan²α+1)=2

有条件可得tana=-2,化简原式得2sin^2a-sinacosa,再化为1-cos2a+1/2sin2a，用万能公式得sin2a=-4/5，cos2a=-3/5，易得原始=6/5.
PS：方法没问题，计算不知道正确与否，你自己验证。

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