Question

19.已知cosa=1/7,cos(a-b)=13/14,0<b<a<圆周率/2，(1)求tan2a的值;(2)求b

Answer 1

根据cosa=1/7，求出sina=（4根号3）/7
tana=4根号3
所以tan2a=-（仔烂8根号3）/47
sin(a-b)=（3根号3）/14
tan(a-b)=（3根号坦戚陵3）/13
tan(a-b)=(tana-tanb)/(1+tanatanb)=(3根号让戚3）/13
求出tanb=根号3
所以b=60°

Answer 2

(1)∵cosa=1/7
∴sina=4√3/和慧7,tana=4√3
tan2a=sin2a/雀裂cos2a=2*sina*cosa/((cosa)^2-(sina)^2)=-8√顷棚闭3/47
(2)sin(a-b)=3√3/14
tan(a-b)=3√3/13
tan(a-b)=(tana-tanb)/(1+tanatanb)
tanb=√3
b=60°

Answer 3

cosa=1/7 sina>蔽明悔0 sina=4/宏正7*根下3
tan2a=2tana/(1-tana^2)= -8/47*根下3
sin(a-b)=3/14 *根下3
tan(a-b)= 3/13*根槐斗下3
tanb=tan(a-(a-b))={tana-tan(a-b)} / (1+tana*tan(a-b))= 根下3
b=60度