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ǰλãҳ>>Ϊʲô-1<=cos(A-B)<=1?>>

Ϊʲô-1<=cos(A-B)<=1?

a*bab a*b=|a|*|b|*cos abļн

cos(a-b/2)=-1/9cos(a/2-b)=2/3 /2

⣺0

֪0

cosa=1/7sina=43/7 tana=sina/cosa = 43 tan(-a)=-tana=- 43 cosb=cos(a-b-a)=cos(a-b)cosa+sin(a-b)sina 0

90

0

|a|=[(1+cosa)^2+(sina)^2] =(2+2cosa)=(4cos(a/2)^2)=2cos(a/2) Ϊ0

θa, |xn-0|=|1/n*cos n/2|

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