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化简:cos((6n+1)/6*π+x)+cos((6n-1)/6*π-x)(n属于z)
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化简:cos((6n+1)/6*π+x)+cos((6n-1)/6*π-x)(n属于z)
▼优质解答
答案和解析
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))
=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))
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