作业帮若f(x)=asin(x+π/4)+bsin(x-π/4)(ab≠0)是偶函数,则有序实数对（a,b）可以是______(只需写一组)
来源：学生作业帮助网 编辑：作业帮 时间：2024/11/25 21:15:54
若f(x)=asin(x+π/4)+bsin(x-π/4)(ab≠0)是偶函数,则有序实数对（a,b）可以是______(只需写一组)
若f(x)=asin(x+π/4)+bsin(x-π/4)(ab≠0)是偶函数,则有序实数对（a,b）可以是______(只需写一组)
若f(x)=asin(x+π/4)+bsin(x-π/4)(ab≠0)是偶函数,则有序实数对（a,b）可以是______(只需写一组)
f(x)是偶函数
也就是f(-x)=f(x)
那么f(-x)=asin(-x+π/4)+bsin(-x-π/4)=f(x)=asin(x+π/4)+bsin(x-π/4)
对比可得到：当a=b时就可以容易得到~ab≠0
f(x)=asin(x+π/4)+bsin(x-π/4)(ab≠0)是偶函数
f(-x)=f(x)
asin(-x+π/4)+bsin(-x-π/4) = asin(x+π/4)+bsin(x-π/4)
a { sin(-x+π/4) - sin(x+π/4) } = b {sin(x-π/4) - sin(-x-π/4) ｝
2acos｛〔(-x+π/4) ＋(...
全部展开
f(x)=asin(x+π/4)+bsin(x-π/4)(ab≠0)是偶函数
f(-x)=f(x)
asin(-x+π/4)+bsin(-x-π/4) = asin(x+π/4)+bsin(x-π/4)
a { sin(-x+π/4) - sin(x+π/4) } = b {sin(x-π/4) - sin(-x-π/4) ｝
2acos｛〔(-x+π/4) ＋(x+π/4)〕／2｝sin｛〔(-x+π/4) －(x+π/4)〕／2｝＝2bcos｛〔(x-π/4) ＋ (-x-π/4)〕／2｝sin｛〔(x-π/4) － (-x-π/4)〕／2｝
2acos｛π/4｝sin｛-x｝＝2bcos｛-π/4｝sin｛x｝
2a×根号2／2×（－sinx）＝2b×根号2／2×sinx
－a＝b
ab≠0
实数对a，b可以是－1,1等
收起