已知tanα+1/tanα=5/2,求2sin2(3π-α)-3cos(π/2+α)sin(3π/2-α)的值

问题描述:

已知tanα+1/tanα=5/2,求2sin2(3π-α)-3cos(π/2+α)sin(3π/2-α)的值

tana+1/tana=5/2
(tan^2a+1)/tana=5/2
2sin2(3π-a)-3cos(π/2+a)sin(3π/2-a)
=-2sin2a-3sinacosa
=-4sinacosa-3sinacosa
=-7sinacosa
=-7sinacosa/(sin^2a+cos^2a)
=-7tana/(tan^2a+1)
=-14/5