sin(A+B)=sinAcosB+cosAsinBcos(A+B)=cosAcosB−sinAsinBtan(A+B)=sin(A+B)cos(A+B)=tanA+tanB1−tanAtanBtan(θ)=sin(θ)cos(θ)=tanθ2+tanθ21−tan2θ2=2tk1−t2k(2tk)2+(1−t2k)2=1解得k=1+t2则sinθ=2t1+t2,cosθ=1−t21+t2sin(A+B)=sinAcosB+cosAsinB\\
cos(A+B)=cosAcosB-sinAsinB\\
tan(A+B)=\frac{sin(A+B)}{cos(A+B)}=\frac{tanA+tanB}{1-tanAtanB}\\
tan(θ)=\frac{sin(θ)}{cos(θ)}=\frac{tan\frac{θ}{2}+tan\frac{θ}{2}}{1-tan^2\frac{θ}{2}}=\frac{\frac{2t}{k}}{\frac{1-t^2}{k}} \\
(\frac{2t}{k})^2+({\frac{1-t^2}{k}})^2=1\\
解得k=1+t^2\\则sinθ=\frac{2t}{1+t^2},cosθ=\frac{1-t^2}{1+t^2}sin(A+B)=sinAcosB+cosAsinBcos(A+B)=cosAcosB−sinAsinBtan(A+B)=cos(A+B)sin(A+B)=1−tanAtanBtanA+tanBtan(θ)=cos(θ)sin(θ)=1−tan22θtan2θ+tan2θ=k1−t2k2t(k2t)2+(k1−t2)2=1解得k=1+t2则sinθ=1+t22t,cosθ=1+t21−t2
从正切公式到三角代换
最新推荐文章于 2025-07-07 15:05:25 发布