求证：[2sin(θ-3π/2)cos(θ+π/2)-1]/1-2sin^2 θ=[tan(9π+θ)+1]/tanθ-1_数学解答

求证：[2sin(θ-3π/2)cos(θ+π/2)-1]/1-2sin^2 θ=[tan(9π+θ)+1]/tanθ-1

人气：458 ℃　时间：2019-10-17 13:54:23

解答

左式=[2sin(θ+p/2)cos(θ+p/2)-1]/[1-2(sinθ)^2]=[sin(p+2θ)-1]/[1-2(sinθ)^2]=-(sin2θ+1)/cos2θ=-(cosθ+sinθ)^2/[(cosθ-sinθ)(cosθ+sinθ)]=-(cosθ+sinθ)/(cosθ-sinθ)=-(1+tanθ)/(1-tanθ)右式=(ta...