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
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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