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化简sin[(4n-1)π/2-a]+cos[(4n+1)π/2-a]
人气:102 ℃ 时间:2020-06-19 08:38:40
解答
sin[(4n-1)π/2-a]+cos[(4n+1)π/2-a]
=sin(2nπ-π/2-a)+cossin(2nπ+π/2-a)
=sin(-π/2-a)+cos(π/2-a)
=-sin(π/2+a)+sina
=-cosa+sina
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