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

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

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

π/4所以π/4所以sin(a-b/2)>0,cos(a/2-b)>0 [sin(a-b/2)]^2=1-[cos(a-b/2)]^2=1-1/81=80/81
所以sin(a-b/2)=4(根号5)/9 同理，
cos(a/2-b)=根号5/3

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π/4所以π/4所以sin(a-b/2)>0,cos(a/2-b)>0 [sin(a-b/2)]^2=1-[cos(a-b/2)]^2=1-1/81=80/81
所以sin(a-b/2)=4(根号5)/9 同理，
cos(a/2-b)=根号5/3
sin(a/2-b)+sin(a-b/2)=2sin[3(a-b)/4]*cos[(a+b)/4] cos(a/2-b)-cos(a-b/2)=2sin[3(a-b)/4]*sin[(a+b)/4]
两式相除得:
tan[(a+b)/4]=[√5/3-(-1/9)]/[2/3+4√5/9] 然后利用万能公式，即可求出cos[(a+b)/2]来

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