Question

求证：(3sin2θ-4cos2θ)/(2tanθ-1)-sin2θ=4cosθ^2

Answer 1

(3sin2θ-4cos2θ)/(2tanθ-1)-sin2θ
=(6sinθcosθ+4sinθ^2-4cosθ^2)/((2sinθ-cosθ)/cosθ)-2sinθcosθ
=2cosθ(2sinθ^2+3sinθcosθ-2cosθ^2)/(2sinθ-cosθ)-2sinθcosθ
=2cosθ(2sinθ-cosθ)(sinθ+2cosθ)(2sinθ-cosθ)-2sinθcosθ
=2sinθcosθ+4cosθ^2-2sinθcosθ
=4cosθ^2

Answer 2

(3sin2θ-4cos2θ)/(2tanθ-1)-sin2θ=4cosθ^2=2+2cos2θ
再运用万能公式sin2α=[2tanα]/{1+[tanα]^2} cos2α=[1-(tanα)^2]/{1+[tanα]^2}
将式子统一化为只含tanθ的式子 ，再验证成立即可