Question

cos1°+cos2°+cos3°....+co180°=?要过程

Answer 1

注意到
cos1+cos179
=cos1+cos(180-1)
=cos1-cos1
=0

一般地，有
cosx+cos(180-x)=0

因此
cos1°+cos2°+cos3°....+cos180°
=(cos1+cos179)+(cos2+cos178)+...+(cos89+cos91)+cos90+cos180
=0+0+...+0+0+(-1)
=-1

Answer 2

cos1°+cos179°=0
cos2°+cos178°=0
...
故原式=cos90°+cos180° = -1

Answer 3

结果-1，这个不是让你算的，你画出cos函数图像，你会发现在0°到180°是关于点（90°，0）对称。cos1°+cos179°=0，cos2°+cos178°=0，……cos89°+cos91°=0，cos90°=0，cos180°=-1从而得出结果

Answer 4

cos1°+cos2°+cos3°....+co180°
＝cos1+cos179+(cos2+cos178)+...(cos89+cos91)+cos180
=cos1-cos1+(cos2-cos2)+...(cos89-cos89)-1
=-1