化简:[sin(2π-2)cos(π+x)tan(π-x)]/[cos(π-x)sin(3π-x)sin...
来源:学生作业帮 编辑:搜搜做题作业网作业帮 分类:数学作业 时间:2024/04/28 23:38:42
化简:[sin(2π-2)cos(π+x)tan(π-x)]/[cos(π-x)sin(3π-x)sin...
化简:
[sin(2π-2)cos(π+x)tan(π-x)]/[cos(π-x)sin(3π-x)sin(-π-x)]
化简:
[sin(2π-2)cos(π+x)tan(π-x)]/[cos(π-x)sin(3π-x)sin(-π-x)]
式子中的sin(2π-2)应该是sin(2π-x)吧!
[sin(2π-x)cos(π+x)tan(π-x)]/[cos(π-x)sin(3π-x)sin(-π-x)]
=[sin(2π-x)cos(π+x)tan(π-x)]/[cos(π-x)sin(3π-x)sin(-π-x)]
=[sin(-x)(-cosx)(-tanx)]/[-cosxsin(π-x)sin(π-x)]
=(-sinx*cosx*tanx)/(-cosx*sinx*sinx)
=tanx/sinx
=1/cosx
[sin(2π-x)cos(π+x)tan(π-x)]/[cos(π-x)sin(3π-x)sin(-π-x)]
=[sin(2π-x)cos(π+x)tan(π-x)]/[cos(π-x)sin(3π-x)sin(-π-x)]
=[sin(-x)(-cosx)(-tanx)]/[-cosxsin(π-x)sin(π-x)]
=(-sinx*cosx*tanx)/(-cosx*sinx*sinx)
=tanx/sinx
=1/cosx
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