已知sin(a-B)＝1求证tan(2a+B)

a+b=(cosa+cosb,sina+sinb)a-b=(cosa-cosb,sina-sinb)(a+b)(a-b)=cosa^2-cosb^2+sina^2-sinb^2=1-1=0所以向量a+

因为cos(a+B)+1=0所以a+B=180度所以sin(2a+B)+sinB=sin(360度-B)+sinB=-sinB+sinB=0

(x/a)cosθ+(y/b)sinθ=1[(x/a)cosθ+(y/b)sinθ]^2=1(x/a)sinθ-(y/b)cosθ=1[(x/a)sinθ-(y/b)cosθ]^2=1[(x/a)co

sin²a/sin²b+cos²acos²c=1sin²a+sin²bcos²acos²c=sin²bsin&

（1）sin(A+B)=3/5,sin(A-B)=1/5sin(a+b)=sinAcosB+sinBcosA=3/5sin(a-b)=sinAcosB-sinBcosA=1/5两式相加相减后可得：si

sin(A+B)=sinAcosB+cosAsinB=3/5sin(A-B)=sinAcosB-cosAsinB=1/5两式分别相加减,得sinAcosB=2/5cosAsinB=1/5两式相除tan

tan（A＋B）＝2tanA,sin(A+B)cosA=2sinAcos(A+B)sin(2A+B)=sin(A+A+B)=sinAcos(A+B)+cosAsin(A+B)=sinAcos(A+B)

tan(A-B)=(tanA-tanB)/(1+tanA*tanB)tan(A-B)/tanA+sin²C/sin²A=1左右移项得1-[(tanA-tanB)/(1+tanA*t

tan(A-B)=(tanA-tanB)/(1+tanA*tanB)tan(A-B)/tanA+sin²C/sin²A=1左右移项得1-[(tanA-tanB)/(1+tanA*t

sin(a+b)=1cos(a+b)=0a+b=2kπ+π/2tan(2a+b)+tanb=tan[(a+b)+b]+tanb=-cotb+tan

sin(a+b)=1,cos(a+b)=0tan(2a+b)+tanb=tan[a+(a+b)]+tanb=[sinacos(a+b)+cosasin(a+b)]/[cosacos(a+b)-sina

tan^2A=2tan^B+1(sinA/cosA)^2=2(sinB/cosB)^2+1sin^2A/cos^2A=2sin^2B/cos^2B+1sin^2A*cos^2B=2sin^2Bcos^

由tan²a=2tan²b+1得sin²a/cos²a=2sin²b/cos²b+1sin²acos²b=2sin

3sinB=sin(2A+B)3sin(A+B-A)=sin(A+b+A)3sin(A+B)cosA-3cos(A+b)sinA=sin(A+B)cosA+sinAcos(a+b)2sin(A+b)c

分析法倒推tanr=-tan(a-r-a)=[tana-tan(a-r)]/[1+tana*tan(a-r)]tana*tanr=[tan^2a-tana*tan(a-r)]/[1+tana*tan(

证明：sin(a+b)=1→cos(a+b)=√[1-sin^2(a+b)]=0→sin(2a+2b)=2*sin(a+b)*cos(a+b)=0→tan(2a+2b)=sin(2a+2b)/cos(

证明：sin^2B=2sin^2A-1得cos^2B=1-sin^2B=1-(2sin^2A-1)=2(1-sin^2A)=2cos^2A于是tan^2A-2tan^B=sin^2A/cos^2A-2

sin(a+b)=sinacosb+cosasinb=1/2sin(a-b)=sinacosb-cosasinb=1/3所以sinacosb=(1/2+1/3)/2=5/12cosasinb=(1/2

sin(A+B)=3/5,sin(A-B)=1/5sin(A+B)=sinAcosB+sinBcosA=3/5sin(A-B)=sinAcosB-sinBcosA=1/5两式相加相减后可得：sinAc

sin(A+B)=sinAcosB＋cosAsinB=3/5（1）sin(A-B)=sinAcosB－cosAsinB=1/5（2）（1）－（2）×3可得2sinAcosB=4osAsinB,两边同时