求证(tan^2a-sin^2a)cot^2a=sin^2a

5sinB=sin(2A+B)=sin(A+B+A)=sin(A+B)cosA+cos(A+B)sinA,sinB=sin(A+B-A)=sin(A+B)cosA-cos(A+B)sinA5sin(A

tan^2a+1=2（tan^2b+1)所以1/cos^2a=2/cos^2b2cos^2a=cos^2b=1-sin^2b1+cos2a=1-sin^2bcos2a+sin^2b=0

tan^2θ+1=2（tan^2a+1）sec^2θ=2sec^2acos^2a=2cos^2θ所以cos2θ=-2sin^2acos2θ+2sin^2a=0

(1＋2sinacosa)/(cos²a－sin²a)=[sin²a＋2sinacosa＋cos²a]/[cos²a－sin²a]=[(si

3sinb=sin(2a+b)可得sin(2a+b)-sinb=2sinb由两角正弦差的公式：sin(2a+b)-sinb=2cos[(2a+b+b)/2]sin[(2a+b-b)/2]=2cos(a

tan(A+B)=2tanAsin(A+B)*cosA=2sinA*cos(A+B)①sin(A+B)*cosA-sinA*cos(A+B)=sinA*cos(A+B)sinB=sinA*cos(A+

cos(a/2)不等于0.tan(a/2)=sin(a/2)/cos(a/2)=2sin(a/2)cos(a/2)/{2[cos(a/2)]^2}=sin(a)/[1+cos(a)]当sin(a/2)

分析法就是从待证的结论推到已知条件,综合法是从结论和条件两个方向出发证明.要证（a^2-b^2)^2=16ab,只需证a^2-b^2=4(ab)^1/2,再用两个条件代进去,就可证明.

第一问的方法是将1拆成sin²a+cos²a,然后就能算了第二问用到常用的倍角公式：cos2θ=cos²a-sin²a=2cos²a-1=1-2sin

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^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

(sin^2A+cos^2A+sinAcosA)/(sin^2A+cos^2A)（分子分母同时除以cos^2A）=(sin^2A/cos^2A+cos^2A/cos^2A+sinAcosA/cos^2

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^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

左边=1/[(sin^2)a(cos^)a]-(cos^2)a/(sin^2)a=[1-(cos^4)a]/[(sin^2)a(cos^2)a]=[1+(cos^2)a][1-(cos^2)a]/[(

cosA-2sin(A-B)=cos[A-B+B]-2sin(A-B)=cos[(A-B)+B]-2sin(A-B)=cos(A-B)cosB-sin(A-B)sinB-2sin(A-B)=0=>co

答:5sinα=3sin(α-2β)5sin[(α-β)+β]=3sin[(α-β)-β]5sin(α-β)coaβ+5cos(α-β)sinβ=3sin(α-β)cosβ-3cos(α-β)sinβ