Question

如何将cos(1/2(a+b))cos(1/2(a-b))变成1/2(cosa+cosb)的形式

Answer 1

解：令 x=(a+b)/2,
y=(a-b)/2,
则 a=x+y,
b=x-y.
所以 cos a=cos (x+y)=cos x cos y -sin x siny,
cos b=cos (x-y)=cos x cos y +sin x siny.
所以 (1/2)(cos a +cos b)=cos x cos y.
即 cos [(a+b)/2] cos [(a-b)/2] = (1/2)(cos a +cos b).

= = = = = = = = =
至于其他的和差化积，积化和差之类的，都可以用这办法做出来。

Answer 2

解答：

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

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

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

所以：cos[(a+b)/2]cos[(a-b)/2]=1/2(cosa+cosb)

Answer 3

公式cosαcosβ = 1/2*[cos(α+β)+cos(α-β)]

cos(1/2(a+b))cos(1/2(a-b))
=1/2*{cos(1/2(a+b)+1/2(a-b)]+cos[1/2(a+b)+1/2(a-b)]}
=1/2(cosa+cosb)

Answer 4

假设A=1/2(a+b),B=1/2(a-b)
A+B=a,A-B=b
cosAcosB
=1/2[cos(A+B)+cos(A-B)]
=1/2(cosa+cosb)