Question

已知函数f(x)=sin(wx+π/4)-asin(wx-π/4)是最小的正周期为的偶函数,求w和a的值

Answer 1

f(x)=sin(wx+π/4)-asin(wx-π/4)
函数f(x)是偶函数
f(-x)=f(x)
sin(-wx+π/4)-asin(-wx-π/4)=sin(wx+π/4)-asin(wx-π/4)
==>-sin(wx-π/4)+asin(wx+π/4)=sin(wx+π/4)-asin(wx-π/4)
(a-1)[sin(wx+π/4)+sin(wx-π/4)]=0
因为sin(wx+π/4)+sin(wx-π/4)是变量
所以a-1=0, a=1
f(x)=sin(wx+π/4)-sin(wx-π/4)
=√2coswx
T=π,2π/w=π==>w=2
∴a=1,w=2

Answer 2

=√2 sin(wx 阿发 45°) =√2 cos(wx 阿发- 45°) （因为和偶函数∵最小正周期为π，所以w=2, 又因为f(x)为偶函数，coswx本身是偶函数