Question

已知sin(3π+θ)=1/4，求cos(π+θ)/cosθ[cos(π+θ)-1]再+cos(θ-2π)/cos(θ+2π)cos(π+θ)+cos(-θ)

求详细解答过程……谢谢……在线等

Answer 1

为输入方便起见
令θ=a,
则sin(3π+θ)=-sinθ=-sina=1/4,sina=-1/4
cos（π+θ）/(cosθ[cos(π+θ)-1])+cos(θ-2π)/(cos(θ+2π)cos(θ+π)+cos(-θ))
=-cosa/[cosa(-cosa-1)]+cosa/[cosa(-cosa)+cosa]
=1/(1+cosa)+1/(1-cosa)
=(1-cosa)+(1+cosa)/(1-cos^2a)
=2/sin^2a
所以，cos（π+θ）/(cosθ[cos(π+θ)-1])+cos(θ-2π)/(cos(θ+2π)cos(θ+π)+cos(-θ))
=2/（1/16）=32