数学三角函数 A，B都是锐角 且A+B=2π/3 则cos(A-B)的取值范围

来源：百度知道 编辑：UC知道 时间：2024/05/10 07:03:03
谢啦 急

A=2π/3-B
A-B=2B-2π/3
0<B<π/2
-2π/3<2B-2π/3<π/3
-1/2=<cos(2B-2/3π)<=1
既-1/2=<cos(A-B）<=1

B=2π/3-A, A-B=A-(2π/3-A)= 2A-2π/3

0<B<π/2, -2π/3<2B-2π/3<π/3

-1/2 <= cos(2B-2/3π) <=1, -1/2 <=cos(A-B）<=1

COSACOSBCOSC
=COSACOSB(-COS(A+B))
=-1/2(COS(A+B)+COS(A-B))COS(A+B)
=-1/2(COS(A+B)COS(A+B)+COS(A-B)COS(A+B))
=-1/2((-COSC)(-COSC)+1/2(COS2A+cos(-2B))
=-1/2(1-SINCSINC+1/2(1-2SINASINA+1-2sinbsinb))
=sinasina+sinbsinb+sincsinc
由于A,B,C都是锐角
所以SINA,SINB,SINC大于SINASINA.......且都为正
所以SINA+SINB+SINC大于COSACOSBCOSC

因为A+B=2π/3，所以有A=2π/3-B ，
从而A-B=2B-2π/3。
又0<B<π/2 ，所以，-2π/3<2B-2π/3<π/3，进而-1/2<cos(2B-2/3π)≤1。
即-1/2<cos(A-B）≤1 。