tan(a b)=2tana 求证

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

准备知识tannπ=0,tan(x+y)=(tanx+tany)/(1-tanxtany)tan(nπ+a)=(tannnπ+tana)/(1-tannnπtana)=(0+tana)/(1-0)=t

(1＋tan^2a)/(1+cot^2a)=[(cos^2a+sin^2a)/cos^2a]/[(sin^2a+cos^2a)/sin^2a]=sin^2a/cos^2a=tan^2a；(1-tana

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+

如图.

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

证明：[(sina+cosa)^2-(sina-cosa)^2]/(tana-sinacosa)=[sina^2+2sinacosa+cosa^2-sina^2+2sinacosa-cosa^2]/(

可能我算错了我算的是tan(A-B)=sin2B/(3-cos2B)下面是我算的tan(A-B)=(tanA-tanB)/(1+tanAtanB)=tanB/[1+(tanB)^2]=sinAcosA

∵tan＝tan=tan=-tana再问：没看明白，能否详细再答：因为tan的奇函数，故：tan(360-a)=-tana

注意到tan(A+B)=tanA+tanB\1-tanAtanB.有左=tan(a+b)-tana\1+tanatan(a+b)=tan(a+b)+tan(-a)\1-tan(-a)tan(a+b)=

左边=(tana-tanb)/(-1/tana+cotb)=(tana-tanb)/(-1/tana+1/tanb)上下乘tanatanb=tanatanb(tana-tanb)/(tana-tanb

tanA=2tan(A/2)/[1-(tan(A/2))^2]-2/tanA=-2*[1-(tan(A/2))^2]/[2tan(A/2)]=[(tan(A/2))^2-1]/(tanA/2)tan(

证明：tan(a+b)=(tana+tanb)/(1-tana·tanb)∴tan(a+π/4)=[tana+tan(π/4)]/[1-tana·tan(π/4)]=(1+tana)/(1-tana)

tan(360°-a）=tan(-a)=-tana

tan(a/2)-1/(tana/2)=sin(a/2)/cos(a/2)-cos(a/2)/sin(a/2)通分=[sin²(a)-cos²(a/2)]/[sin(a/2)cos

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

7sinB=sin(2A+B)7sin(A+B-A)=sin(A+B+A)7sin(A+B)cosA-7sinAcos(A+B)=sin(A+B)cosA+sinAcos(A+B)6sin(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(

这个本来就是公式推公式sin(A-B)=sinAcosB-cosAsinBcos(A-B)=cosAcosB+sinAsinBtan(A-B)=(sinAcosB-cosAsinB)/(cosAcos