Question

cos(2 兀/5)cos(兀/5-2)的值详细过程＿＿＿还有cosa=1/2其中a属于(-兀/2,0) sin(a/2)=?

天阿你们都错啦，是cos(2加兀/5)cos(兀/5-2)

Answer 1

应该是cos(2 π/5)cos(π/5-2π）
所以cos(2 π/5)cos(π/5-2π)
=cos(2 π/5)cos(π/5)
=[2sin(π/5)cos(π/5)cos(2π/5)]/[2sin(π/5)]
=sin(2π/5)cos(2π/5)/[2sin(π/5)]
=2sin(2π/5)cos(2π/5)/[4sin(π/5)]
=sin(4π/5)/[4sin(π/5)]
=sin(π/5)/[4sin(π/5)]
=1/4
a∈(-π/2,0)，∴a/2∈(-π/4,0)，∴sin(a/2)<0
∴sin(a/2)=-√[(1-cosa)/2]=-√(1/4)=-1/2

Answer 2

cos(兀/5)cos(2兀/5)=2cos(兀/5)sin（π/5）cos(2兀/5)/2sinπ/5

= sin（2π/5）cos2π/5/2sinπ/5

=sin4π/5 / 4sinπ/5

=sinπ/5 / 4sinπ/5

=1/4

-Pai/2<a<0,则有-Pai/4<a/2<0.sina/2<0
cosa=1-2(sina/2)^2
1/2=1-2(sina/2)^2
(sina/2)^2=1/4
sina/2=-1/2.

Answer 3

不是楼上的有错是cos(2 兀/5)cos(兀/5-2)