如题所述
tanb=1/2,
tan2b=2tanb/(1-tan²b)=1/(1-1/4)=4/3
æ以
tan(a+2b)=(tana+tan2b)/(1-tanatan2b)
=(7+4/3)/(1-7Ã(4/3))
=(25/3)/(-25/3)
=-1
å 为a,bå为éè§
æ以
a+2b=3Ï/4
tan2b=2tanb/(1-tan²b)=1/(1-1/4)=4/3
æ以
tan(a+2b)=(tana+tan2b)/(1-tanatan2b)
=(7+4/3)/(1-7Ã(4/3))
=(25/3)/(-25/3)
=-1
å 为a,bå为éè§
æ以
a+2b=3Ï/4
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第1个回答 2012-04-27
tan2b=2tanb/1-tan^2 b=1/(1/4)=4
tan(a+2b)=(tana+tan2b)/(1-tanatan2b)=(7+4)/(1-28)=-11/27.
a+2b=arc tan (-11/27).
a,b均为锐角
0<a<π/2,0<b<π/2,
0<a+2b<3π/2.
a+2b=arc tan (-11/27).
tan(a+2b)=(tana+tan2b)/(1-tanatan2b)=(7+4)/(1-28)=-11/27.
a+2b=arc tan (-11/27).
a,b均为锐角
0<a<π/2,0<b<π/2,
0<a+2b<3π/2.
a+2b=arc tan (-11/27).
第2个回答 2012-04-27
tan2b=2tanb/(1-(tanb)^2)=2*(1/2)/(1-1/4)=4/3
tan(a+2b)=(tana+tan2b)/(1-tanatan2b)=(7+4/3)/(1-7*4/3)=(25/3)/(-25/3)=-1
因为a,b都是锐角,则有0<=a+2b<=270度,
所以有:a+2b=135度,
tan(a+2b)=(tana+tan2b)/(1-tanatan2b)=(7+4/3)/(1-7*4/3)=(25/3)/(-25/3)=-1
因为a,b都是锐角,则有0<=a+2b<=270度,
所以有:a+2b=135度,
