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:2024/11/09 09:57:15

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