题目

求一道三角函数题计算sin^2 1°+sin^2 2°+sin^2 3°+……sin^2 88°+sin^2 89°谢谢了,

提问时间：2021-03-16

答案

sin^2 1°+sin^2 2°+sin^2 3°+……sin^2 88°+sin^2 89° =(sin^2 1°+sin^2 89°）+（sin^2 2°+sin^2 88°）+……+（sin^2 44°+sin^2 46°）+sin^2 45° =(sin^2 1°+cos^2 1°）+（sin^2 2°+cos^2 2°）+……+（sin^2 44°+cos^2 44°）+sin^2 45° =1+1+……+1+1/2 =44+1/2 =89/2