Question

cosa=1/7.cos(a-b)=13/14. 0<b<a<兀/2 (1)求tan2a的值 (2)求b的大小 要过程

Answer 1

cos(a-b-a)=cosb=cos(a-b)cosa+sin(a-b)sina
0<b<a<π/2,0<a-b<π/2
sina=4√3/7 sin(a-b)=3√3/14
所以cos(a-b-a)=cosb=cos(a-b)cosa+sin(a-b)sina
=13/14 * 1/7 +4√3/7 * 3√3/14
=13/98 + 36/98
=49/98
=1/2
b=π/3
tana=sina/cosa = 4√3
tan2a=2tana/[1-(tana)^2]=-8√3/47

Answer 2

解：
已知cosa=1/7,cos(a-b)=13/14

0＜b＜a＜π/2
则0＜a-b＜π/2

所以sina=√[1-(1/7)²]=4√3/7
sin(a-b)=√[1-(13/14)²]=3√3/14

所以tan2a=sin2a/cos2a=2sinacosa/(cos²a-sin²a)
=2*(4√3/7)*(1/7)/[(1/7)²-(4√3/7)²]
=-8√3/47

cosb=cos[a-(a-b)]=cosacos(a-b)+sinasin(a-b)
=(1/7)*(13/14)+(4√3/7)*(3√3/14)
=1/2
所以b=π/3