f(x)=lg(sinx+根号下(1+sin^2x).
f(-x)
=lg[-sinx+√(1+sin²x)]
=lg{[-sin²+(1+sin²x)]/[sinx+√(1+sin²x)]}
这步是为什么?
人气:311 ℃ 时间:2019-09-25 09:39:13
解答
f(-x)=lg(sin(-x)+根号1+sin²(-x))=lg(-sinx+根号1+sin²x)真数分子分母同乘(sinx+√(1+sin²x))=lg{[-sin²x+(1+sin²x)]/[sinx+√(1+sin²)]}=lg1/(sinx+√(1+sin²))=-lg(sinx+√(1...
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