已知psina-qcosa=根号下p方 q方，求(pcosa-2qsina)/(3pcosa-4qsina)已知psina-qcosa=根号下p方+q方，求(pcosa-2qsina)/(3pcosa-4qsina
热心网友
已知psina-qcosa=根号下p方+q方，求(pcosa-2qsina)/(3pcosa-4qsina 因psina-qcosa=√(p^2+q^2) ,所以两边平方得：Pcosa +qsina =0所以pcosa =-qsina 所以原式＝(-qsina-2qsina)/(-3qsina -4qsina)＝3/7
热心网友
psinA-qcosA=(p^2+q^2)---p^2*(sinA)^2-2pqsinAcosA+q^2*(cosA)^2=p^2+q^2---p^2*[1-(sinA)^2]+2pqsinAcosA+q^2*[1-(cosA)^2]=0---p^2*(cosA)^2+2*pcosA*qsinA+q^2*(sinA)^2=0---(pcosA+qsinA)^2=0---pcosA+qsinA=0---pcosA=-qsinA---(pcosA-2qsinA)/(3pcosA-4qsiA)=(-qsinA-2qsinA)/(-3qsinA-4qsinA)=(-3qsinA)/(-7qsinA)=3/7
热心网友
解：由psina-qcosa=根号下p方+q方及辅助角公式得： psina-qcosa=[根号下p方+q方]sin(a+&),其中sin&=-q/[根号下p方+q方]; cos&=p/[根号下p方+q方]; 据已知有sin(a+&)=1,a+&=2kπ+π/2.&=2kπ+π/2-a,sin&=cosa,cos&=sina; tana=cot&=-p/q; (pcosa-2qsina)/(3pcosa-4qsina )=(p-2q/tana)/(3p-4qtana) =(p+2q^2/p)/(3p+4p)=1/7+2q^2/(7p^2).