Tính giá trị các biểu thức:
a) $A=\cos 20^0+\cos 40^0+\cos 60^0+...+\cos 140^0+\cos 160^0+\cos 180^0   $
b) $B=\sin 20^0+\sin 40^0+\sin 60^0+...+\sin 320^0+\sin 340^0+\sin 360^0    $
a) Ta có: $A=(\cos 20^0+\cos 160^0 )+(\cos 40^0+\cos 140^0 )+...+(\cos 80^0+\cos 100^0 )$
           $+\cos 180^0 $

Các tổng trong dấu ngoặc là các góc bù nhau :  $20^0+160^0=180^0,...$, nên các cosin của chúng đối nhau và các tổng trong dấu ngoặc bằng  $0$.
Vậy   $A=\cos 180^0=-1 $.

b) Tương tự ta có:  $B=0$.

Thẻ

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