已知sin(a+Kπ)=-2cos(a+Kπ),K∈Z则 4sina-2coxa/5cosa+3sina

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已知sin(a+Kπ)=-2cos(a+Kπ),K∈Z则 4sina-2coxa/5cosa+3sina
已知sin(a+Kπ)=-2cos(a+Kπ),K∈Z则 4sina-2coxa/5cosa+3sina

已知sin(a+Kπ)=-2cos(a+Kπ),K∈Z则 4sina-2coxa/5cosa+3sina
4sina-2coxa/5cosa+3sina
=4tana-2 / 5+3tana ①
sin(a+Kπ)=-2cos(a+Kπ)
sin(a+kπ)/cos(a+kπ)=-2
tan(a+kπ)=-2
因为k为整数,不管k为奇数还是偶数,都有tan(a+kπ)=tana=-2
将tana=-2代入①式得
4sina-2coxa/5cosa+3sina =-5

若k为偶数,条件等价为sina=-2cosa 即tana=-2 若k为奇数,条件等价为-sina=2cosa 即tana=-2 所以tana=-2 (4sina-2cosa)/(5cosa 3sina) (上下同时