Question

已知cos(a-B/2)=-1/9,sin(a/2-B)=2/3,且π/2<a<π,0<B<π/2,求cos(a+B）的值

Answer 1

∵π/2<a<π，0<b<π/2,
∴π/4<a-b/2<π,-π/4<a/2-b<π/2.
∴sin[a-(b/2)]>0,cos[(a/2)-b]>0.
∵cos[a-(b/2)]=-1/9,sin[(a/2)-b]=2/3，
∴sin[a-(b/2)]=4√5/9,cos[(a/2)-b]=√5/3.
故cos(a+b)=2cos²[(a+b)/2]-1
=2cos²[(a-b/2）-(a/2-b)]-1
=2[cos(a-b/2)cos(a/2-b)+sin(a-b/2)sin(a/2-b)]²-1
=2[(-1/9)(√5/3)+(4√5/9)(2/3)]²-1
=-239/729.

Answer 2

π/4<a-B/2<π, -π/4<a/2-B<π/2
所以：sin(a-B/2)=4√5/9 ，cos(a/2-B)=√5/3
cos[(a-B/2)-(a/2-B)]=cos[(a+b)/2]=cos(a-B/2)cos(a/2-B)+sin(a-B/2)sins(a/2-B)=-1/9 * √5/3+
4√5/9 * 2/3=7√5/27
cos(a+B)=2cos[(a+b)/2]^2-1=-239/729