求cos1°+cos2°+cos3°+...+cos178°+cos179°的值
用合并法~~~
求cos1°+cos2°+cos3°+...+cos178°+cos179°的值（这里的数字都是度数比如1°）
答案:3 悬赏:20
解决时间 2021-02-19 06:34
- 提问者网友：空白
- 2021-02-19 03:24
最佳答案
- 二级知识专家网友：荒唐后生
- 2021-02-19 04:53
cos1°+cos2°+cos3°+...+cos178°+cos179°
=（cos1°+cos179°）+（cos2°+cos178°）+……+（cos89°+cos91°）+cos90°
=（cos1°-cos1°）+（cos2°-cos2°）+……+（cos89°-cos89°）+cos90°
=cos90°
=0
=（cos1°+cos179°）+（cos2°+cos178°）+……+（cos89°+cos91°）+cos90°
=（cos1°-cos1°）+（cos2°-cos2°）+……+（cos89°-cos89°）+cos90°
=cos90°
=0
全部回答
- 1楼网友：许你一世温柔
- 2021-02-19 07:34
2cos1度 可以首相和末尾相加就像从1加到100一样
- 2楼网友：温柔刺客
- 2021-02-19 06:28
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
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