已知sinπ-a=log8

很简单前边那个先把对数化了是-2/3后边是tan（a）=-2根号5/5

log8(a)+log4(b^2)=5===>1/3*log2(a)+log2(b)=5---------1)log8(b)+log4(a^2)=7===>1/3*log2(b)+log2(a)=7-

log(8)3=lg3/3lg2=a,log(3)5=lg5/lg3=b,lg2+lg5=1lg5=3ab/(1+3ab)

log89=lg9/lg8=2lg3/3lg2=a;log320=lg20/lg3=(1+lg2)/lg3=b;从此建立二元一次方程组,解得lg2=2/(3ab-2),

a=lg9/lg8=2lg3/3lg2lg2=2lg3/(3a)b=lg5/lg2=(1-lg2)/lg2=1/lg2-1b+1=1/lg2=3a/(2lg3)2lg3=3a/(b+1)lg3=3a/

由题意可知：a>0且b>0因为½log8（a）+log4（b）=5/2,所以：log8（a）+2log4（b）=5（1）又log8(b)+log4(a^2)=7,则：log8(b)+2log

a=log4(3)=lg3/lg4所以lg3=alg4=2alg2log8(9)=lg9/lg8=2lg3/3lg2=4alg2/3lg2=4a/3log4(3)=lg3/2lg2=(1/2)*log

sina=-2/3,因为a∈（-π/2,0）,所以cosa=√5/3cos(π+a)=-cosa=-√5/3

sin(pai-a)=sina=lg(1/4)/lg8sina=-2lg2/3lg2=-2/3a在第四象限则cosa>0sin²a+cos²a=1所以cosa=√5/3所以原式=-

sin(3/2pai-A)=sin(3/2pai-A-2pai)=sin(-1/2pai-A)=-sin(1/2pai+A)=-(-cosA)=cosA=log8(1/4)=log2^3(2^-2)=

由sin(π-α)=log81/4,得：sinα=lg(1/4)/lg8=(-2lg2)/(3lg2)=-2/3由因为：(sinα)^2+(cosα)^2=1且α∈(-π/2,0)所以：cosα=√5

如果8,4是底数第一个方程等价于1/6log2（a）+1/2log2(b)=2.5(1)第二个等价于log2(a)+1/3log2(b)=7(2)(1)*2-(2)*31/3log2(a)-3log2

a=log89=lg9/lg8=2lg3/3lg2b=log320=lg20/lg3=(1+lg2)/lg3ab=2(1+lg2)/3lg2=(2/3lg2)+1/3ab-1/3=2/3lg2lg2=

见图

log83=a==>8^a=3log35=b==>3^b=58^(ab)=52^(3ab)=5∴log25=3ab2^(3ab+1)=10所以log210=3ab+1∵lg5=log105=(log2

log8^3=a/3,log8^9=2a/3

log8四分之一=-2lg2/3lg2=-2/3sin（π-α）=log8¼=-2/3,sina=-2/3cosa=√5/3tan（2/3π+α）=(sin2/3πcosa+cos2/3

先介绍换底公式：log(a)b=lgb/lga证明：设log(a)b=t则a^t=b,两边取以10为底的对数lga^t=lgbtlga=lgb所以t=lgb/lga所以log(a)b=lgb/lga＝

sin(π-a)=log8(1/4)={log2(1/4)}/{log2(8)=(-2)/3=-2/3sina=-2/3a属于（-π/2,0）tan(3π/2+a)=tan{2π-π/2+a)=tan