已知[cos(a-b)]^2-[cos(a+b)]^2=1/2,(1+cos2a)(1+cos2b)=1/3求证[tan（a+b)]^2=47还有一条 tan(a/2)=2 求（1+cosa+cos2a+cos3a)/(1-cosa-2(cosa)^2)
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已知[cos(a-b)]^2-[cos(a+b)]^2=1/2,(1+cos2a)(1+cos2b)=1/3求证[tan（a+b)]^2=47还有一条 tan(a/2)=2 求（1+cosa+cos2a+cos3a)/(1-cosa-2(cosa)^2)
已知[cos(a-b)]^2-[cos(a+b)]^2=1/2,(1+cos2a)(1+cos2b)=1/3
求证[tan（a+b)]^2=47
还有一条 tan(a/2)=2 求（1+cosa+cos2a+cos3a)/(1-cosa-2(cosa)^2)
已知[cos(a-b)]^2-[cos(a+b)]^2=1/2,(1+cos2a)(1+cos2b)=1/3求证[tan（a+b)]^2=47还有一条 tan(a/2)=2 求（1+cosa+cos2a+cos3a)/(1-cosa-2(cosa)^2)
1.由[cos(a-b)]^ 2-[cos(a+b)]^ 2=1/2得cos(2a-2b)+1-1-cos(2a+2b)=1,
即cos2acos2b+sin2asin2b-cos2acos2b+sin2asin2b=1,
sinasinbcosacosb=1/8…………①
由(1+cos2a)(1+cos2b)=1/3得2(cosa)^2×2(cosb)^2=1/3,
即(cosacosb)^2=1/12…………②
①÷②,得tanatanb=3/2.
由②得(seca)^2*(secb)^2=12,
即[1+(tana)^2][1+(tanb)^2]=12,
1+(tana)^2+(tanb)^2+(tanatanb)^2=12,
(tana)^2+(tanb)^2=12-1-(3/2)^2=35/4.
所以,[tan(a+b)]^2=[(tana+tanb)/(1-tanatanb)]^2
=[(tana)^2+(tanb)^2+2tanatanb]/(1-tanatanb)^2
=[35/4+2×(3/2)]/(1-3/2)^2
=47
2.
tan(a/2)=2
sin(a/2)/cos(a/2)=2
[2sin(a/2)cos(a/2)]/[2cos^2(a/2)]=2
sin(a)/[2cos^2(a/2)-1+1]=2
sin(a)/[1+cos(a)]=2
sin(a)=2+2cos(a)
cos(a)=1/2sin(a)-1
又sin^2(a)+cos^2(a)=1,tan(a/2)=2>0
则有:sin(a)>0
则:sin(a)=4/5,cos(a)=-3/5
[1+cosa+cos2a+cos3a)/[1-cosa-2cos^2 a]
=[1+cosa+(2cos^2 a-1)+(4cos^3 a-3cosa)]/[1-cosa-2cos^2 a]
=[4cos^3 a+2cos^2 a-2cosa]/[(2cosa+1)(1-cosa)]
=[2cosa(2cosa-1)(cosa+1)]/[(2cosa+1)(1-cosa)]
=-33/10
47.