化简[sin(2π-α)cos(π+α)cos(π/2+α)cos(11π/2-α)]/[cos(π-α)sin(π+α)sin(-π-α)sin[(...化简[sin(2π-α)cos(π+α)cos(π/2+α)cos(11π/2-α)]/[cos(π-α)sin(π+α)sin(-π-α)sin[(9π/2)+α]]

来源:学生作业帮助网 编辑:六六作业网 时间:2025/02/03 10:17:44
化简[sin(2π-α)cos(π+α)cos(π/2+α)cos(11π/2-α)]/[cos(π-α)sin(π+α)sin(-π-α)sin[(...化简[sin(2π-α)cos(π+α)co

化简[sin(2π-α)cos(π+α)cos(π/2+α)cos(11π/2-α)]/[cos(π-α)sin(π+α)sin(-π-α)sin[(...化简[sin(2π-α)cos(π+α)cos(π/2+α)cos(11π/2-α)]/[cos(π-α)sin(π+α)sin(-π-α)sin[(9π/2)+α]]
化简[sin(2π-α)cos(π+α)cos(π/2+α)cos(11π/2-α)]/[cos(π-α)sin(π+α)sin(-π-α)sin[(...
化简[sin(2π-α)cos(π+α)cos(π/2+α)cos(11π/2-α)]/[cos(π-α)sin(π+α)sin(-π-α)sin[(9π/2)+α]]

化简[sin(2π-α)cos(π+α)cos(π/2+α)cos(11π/2-α)]/[cos(π-α)sin(π+α)sin(-π-α)sin[(...化简[sin(2π-α)cos(π+α)cos(π/2+α)cos(11π/2-α)]/[cos(π-α)sin(π+α)sin(-π-α)sin[(9π/2)+α]]
-sinα*-cosα*-sinα*sinα/-cosα*-sinα*sinα*-cosα
=tanα

只是需要答案哈?不要过程吧

+_+ 不懂