Question

cos(a-b/2)= - 4/5,且 sin(a/2-b)=12/13,且π/2<a<π，0<b<π/2，求cos[(a+b)/2]

Answer 1

∵π/2 < a < π
∴π/4 < a/2 < π/2
同理，0 < b/2 < π/4
则有 π/2 < (a - b/2) < 3π/4
∴sin(a - b/2) =√1 - [cos(a - b/2)]^2 = 3/5
同理，cos(a/2 - b) = 5/13
cos[(a+b)/2] =cos[(a/2)+(b/2)] = cos[(a - a/2) + (b - b/2)] = cos[(a - b/2) - (a/2 - b)]
=cos(a - b/2)*cos(a/2 - b) + sin(a - b/2)*sin(a/2 - b)
=(-4/5)*(5/13) + (12/13)*(3/5)
=16/65

Answer 2

由于π/2 < a < π，0<b<π/2
所以π/4 < a/2 < π/2，0 < b/2 < π/4
所以π/4<a-b/2<π
cos(a-b/2)= - 4/5,
所以sin（a-b/2）=3/5
-π/4<a/2-b< π/4
sin(a/2-b)=12/13
所以
cos（a/2-b）=5/13
cos[(a+b)/2]
=cos[(a - b/2) - (a/2 - b)]
=cos(a - b/2)*cos(a/2 - b) + sin(a - b/2)*sin(a/2 - b)
=(-4/5)*(5/13) + (12/13)*(3/5)
=16/65

Answer 3

你好
已知π/2<a<π，0<b<π/2,cos(a-b/2)= - 4/5,sin(a/2-b)=12/13所以
sin(a-b/2)= 3/5,cos(a/2-b)=5/13
cos[(a+b)/2]
=cos[(a-b/2)-(a/2-b) ]
=cos(a-b/2)cos(a/2-b)+sin(a-b/2)sin(a/2-b)
=-4/13+36/65
=16/65