求cos1°+cos2°+cos3°+...+cos178°+cos179°的值（这里的数字都是度数比如1°）求cos1°+cos2°+cos3°+...+cos178°+cos179°的值

答案为0：
cosα+cosβ=2cos[（α+β）/2] cos[（α-β）/2]
cos1°+cos179°=cos90° cos89°=0
cos2°+cos178°=cos90° cos88°=0
cos3°+cos177°=cos90° cos87°=0
..........
cos89°+cos91°=cos90° cos1°=0
cos90°=0
故cos1°+cos2°+cos3°+...+cos178°+cos179°=0

2cos1度 可以首相和末尾相加就像从1加到100一样

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