Question

求解这道三角函数题的解,高一的要过程啊,谢谢了已知COS（α－β/2）＝－1/9,SIN（α/2－β）＝2/3且90°＜

Answer 1

cos[(a+b)/2]

=cos([a-(b/2)]+[b-(a/2])

=cos(a-(b/2))cos(b-(a/2))-sin(a-(b/2))sin(b-(a/2))

由a,b的取值范围,

sin(a-b/2)＞0

cos(a/2-b)＞0

∴sin(a-b/2)=√(1-(1/81))=4√5/9

cos(a/2-b)=√(1-(4/9))=√5/3

∴原式=-(√5)/3