Question

一一ㄑ

！！

√29
察
14. 試求下列之值
(1)cos20° + cos40° + cos60° + ... + cos340° =
200+ sin 2179° + sin²180°
0
=

Answer 1

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qn

因為 cos(180°-θ) = - cosθ
所以 cos20° + cos160° = cos40° + cos140°
= cos60° + cos120° = cos80° + cos100° = 0

cos180° + cos200° + cos220° + ⋯ + cos340°
= -1 + (-cos20°) + (-cos40°) + ⋯ + (-cos160°)
(∵cos(180°+θ) = - cosθ)

跟前面同理，從20°到160°都會消掉
最後剩下 -1