已知a, b均为锐角,且tana=7,cosb =2 5 求a 2
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cos(a+b)=-5/13,=>sin(a+b)=12/13,=>tan(a+b)=-12/5,tana=3/4,=>tan(a+b)=(tana+tanb)/(1-tana*tanb)=(3/4+
sinA=13k=>cosA=sqrt(1-169k^2)sinB=17k=>cosB=sqrt(1-289k^2)17tanA+13cosB=17=>17*13k/sqrt(1-169k^2)+13
由(sina-cosa)/(sina+2cosa)则:上下同时除以cosa得:(sina/cosa-1)/(sina/cosa+2)=(tana-1)/(tana+2)又tana=3则:(sina-c
A,B为锐角,故A+B∈(0,π),sin(A+B)>0结合cos(A+B)=-11/14,解得sin(A+B)=5√3/14所以tan(A+B)=-5√3/11=(tanA+tanB)/(1-tan
tanA=tan[(B+A)-B]=[tan(B+A)-tanB]/[1+tan2BtanB]=tanB/[1+2(tanB)^2]=1/[1/tanB+2tanB]≤1/(2√2)=√2/4
cosAcosB-sinAsinB=sinAcosB-cosAsinBcosA(sinB+cosB)=sinA(sinB+cosB)因为B是锐角,所以sinB+cosB不等于0cosA=sinAtan
移项!sinA-cosA/sinA+cosA=1/33sinA-3cosA=sinA+cosA2sinA=4cosA∵tanA=sinA/cosA∴tanA=2
答案:1,由(1+tanA)(1+tanB)=2tanA+tanB+tanAtanB=1经通分后可以得到sinAsinB+sinAcosB=cosAcosB-sinAsinBsin(A+B)=cos(
1.cos(A+B)=cosAcosB-sinAsinBsin(A-B)=sinAcosB-sinBcosAcosAcosB-sinAsinB=sinAcosB-sinBcosAcosA(cosB+s
楼主,第(2)应该是tanA+tanB2吧(不过,要证明cosA+cosB>1也可以)举例:tan46°+tan43°≈1.968>√2cos46°+cos43°≈1.426>1证明:(1)∵A、B为
tana=sina/cosa=1/2cosa=2sinasin^2a+cos^2a=1sin^2a+4sin^2a=1sin^2a=1/5sina=根号5/5cosa=2根号5/5
老师不讲或没讲到的题目就不该出这是误区,因为你们老师不是命题人!对于该题记住:sina/cosa=tana^o^
a为锐角,且tana=2sina=2√5/5cosa=√5/5(sina-2)/(2cosa+sina)=(2√5/5-2)/(2√5/5+2√5/5)=(1-√5)/2
cos(A+B)=-11/14,sin(A+B)=√[1-cos^2(A+B)]=5√3/14.tan(A+B)=sin(A+B)/cos(A+B)=-5√3/11.tanB=tan[(A+B)-A]
sina=4√3/7,cosa=1/7sin(a+b)=5√3/14cosb=cos[(a+b)-a]=.
等号两边拆开移项和并同类项约分得sinA=cosA所以tanA=1
1、tanAtanB=tanA+tanB+1tanAtanB-1=tanA+tanB则:tan(A+B)=[tanA+tanB]/[1-tanAtanB]=-1因为A、B为锐角,则:A+B=3π/4,
sinA=12/13tanA=12/5
SinA/SinB=Cos(A+B)SinA=Cos(A+B)SinB=1/2[sin(2B+A)-sinA]3sinA=sin(2B+A)可见当sin(2B+A)=1=3sinA时sinA有最大值1
是(sina+cosa)/(sina-cosa)还是sina+cosa/sina-cosa无括号?是(sina+cosa)/(sina-cosa)的话=[(sina+cosa)/cosa]/[(sin