s i n ( A + B ) = s i n A c o s B + c o s A s i n B c o s ( A + B ) = c o s A c o s B − s i n A s i n B t a n ( A + B ) = s i n ( A + B ) c o s ( A + B ) = t a n A + t a n B 1 − t a n A t a n B t a n ( θ ) = s i n ( θ ) c o s ( θ ) = t a n θ 2 + t a n θ 2 1 − t a n 2 θ 2 = 2 t k 1 − t 2 k ( 2 t k ) 2 + ( 1 − t 2 k ) 2 = 1 解 得 k = 1 + t 2 则 s i n θ = 2 t 1 + t 2 , c o s θ = 1 − t 2 1 + t 2 sin(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
