化简:cos((6n+1)/6*π+x)+cos((6n-1)/6*π-x)(n属于z)
来源:学生作业帮助网 编辑:作业帮 时间:2024/11/17 22:43:05
化简:cos((6n+1)/6*π+x)+cos((6n-1)/6*π-x)(n属于z)
化简:cos((6n+1)/6*π+x)+cos((6n-1)/6*π-x)(n属于z)
化简:cos((6n+1)/6*π+x)+cos((6n-1)/6*π-x)(n属于z)
cos((6n+1)/6*π+x)+cos((6n-1)/6*π-x)
=cos(nπ+π/6+x)+cos(nπ-π/6-x)
=cos(nπ)cos(π/6+x)-sin(nπ)sin(π/6+x)+cos(nπ)cos(-π/6-x)-sin(nπ)sin(-π/6-x)
=cos(nπ)cos(π/6+x)+cos(nπ)cos(-π/6-x)
=cos(nπ)cos(π/6+x)+cos(nπ)cos(π/6+x)
=2cos(nπ)cos(π/6+x)
=2cos(nπ)(cos(π/6)cos(x)-sin(π/6)sin(x))
=2cos(nπ)(cos(x)*(√3)/2-sin(x)/2)
=cos(nπ)(cos(x)*(√3)-sin(x))
=±(cos(x)*(√3)-sin(x))