Question

已知sin(a-b)=1/3,sinacosb=1/6,求cos(2a+2b)

Answer 1

利用两角和差公式计算之:

供参考，请笑纳。

Answer 2

sinacosb = 1/6,
sin(a-b) = sinacosb - cosasinb = 1/6 - cosasinb = 1/3, cosasinb = -1/6
sin(a+b) = sinacosb + cosasinb = 1/6 - 1/6 = 0
cos(2a+2b) = cos2(a+b) = 1 - 2[sin(a+b)]^2 = 1