Question

已知cos(a+π/6）=三分之根号三，求cos(π/6-a）的值

Answer 1

tana=-根号2
则(tana)^2=2
1
(tana)^2=3=1/(cosa)^2
于是(cosa)^2=1/3
(sina)^2=2/3
a∈（π/2，π），在第二象限，则cosa为负，sina为正
于是cosa=-(根号3)/3，sina=(根号6)/3
于是cos(a-π/3)=cosacosπ/3
sinasinaπ/3
=[-(根号3)/3]*(1/2)
[(根号6)/3]*[(根号3)/2]
=(3根号2-根号3)/6
1
(tana)^2=1
[(sina)^2/(cosa)^2]
[tana=sina/cosa]
=[(cosa)^2
(sina)^2]/(cosa)^2
[通分]
=1/(cosa)^2
=[(cosa)^2
(sina)^2=1]
既然(tana)^2=2
那么1
(tana)^2=3
也就是1/(cosa)^2=3
于是(cosa)^2=1/3
既然(cosa)^2
(sina)^2=1
那么(sina)^2=1-(cosa)^2=1-1/3=2/3

Answer 2

cos(π/6-a)
=cos(a-π/6)
=cos(a+π/6-π/3)
=cos(a+π/6)cosπ/3+sin(a+π/6)sinπ/3
=√3/3*1/2±√6/3*√3/2
=√3/6±√2/2