Question

12/13cosa-5/13sina=3/5,sina^2+cosa^2=1这个方程组怎么解

Answer 1

记 b = arctan(5/12),

12/13cosa-5/13sina

= (cosb)(cosa) - (sinb)(sina)

= cos(b + a) = 3/5

b + a = arccos(3/5) + 2kPI,或者

b + a = 2kPI - arccos(3/5).

所以,

a = arccos(3/5) - arctan(5/12) + 2kPI,或者

a = 2kPI - arccos(3/5) - arctan(5/12).

k = 0,1,-1,2,-2,...