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arcsin[sin(19Π/12)]求值
Π为3.1415926
人气:157 ℃ 时间:2020-06-09 16:43:13
解答
y=arcsin(x),定义域[-1,1] ,值域[-π/2,π/2]
arcsin[sin(19Π/12)]=arcsin[sin(2π - 5π/12)]=arcsin[sin(-5π/12)]= -5π/12
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