sin(a+β)cos(r-β)-cos(β+a)sin(β-r)跟tan5π/4+tan5π/12 / 1-tan5π
来源:学生作业帮 编辑:作业帮 分类:数学作业 时间:2024/05/25 20:21:18
sin(a+β)cos(r-β)-cos(β+a)sin(β-r)跟tan5π/4+tan5π/12 / 1-tan5π/12要怎么算
sin(a+β)cos(r-β)-cos(β+a)sin(β-r)
=sin(a+β)cos(r-β)-cos(a+β)sin[-(r-β)]
=sin(a+β)cos(r-β)+cos(a+β)sin(r-β)
=sin[(a+β)+(r-β)]
=sin(a+r)
[(tan5π/4)+(tan5π/12)]/[1-(tan5π/12)]
=[(tan(π+π/4))+(tan5π/12)]/[1-(tan5π/12)]
=[(tan5π/12)+(tanπ/4)]/[1-(tan5π/12)]
=[(tan5π/12)+(tanπ/4)]/[1-(tanπ/4)(tan5π/12)]
=tan(5π/12+π/4)
=tan(2π/3)=-√3
=sin(a+β)cos(r-β)-cos(a+β)sin[-(r-β)]
=sin(a+β)cos(r-β)+cos(a+β)sin(r-β)
=sin[(a+β)+(r-β)]
=sin(a+r)
[(tan5π/4)+(tan5π/12)]/[1-(tan5π/12)]
=[(tan(π+π/4))+(tan5π/12)]/[1-(tan5π/12)]
=[(tan5π/12)+(tanπ/4)]/[1-(tan5π/12)]
=[(tan5π/12)+(tanπ/4)]/[1-(tanπ/4)(tan5π/12)]
=tan(5π/12+π/4)
=tan(2π/3)=-√3
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