已知cos[a-(b/2)]=-1/9,sin[(a/2)-b]=2/3.且π/2

Question

已知cos[a-(b/2)]=-1/9,sin[(a/2)-b]=2/3.且π/2RT

Answer 1

∵π/2∴π/4∴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.