函数y=√2 sin(2x-π)cos【2(x +π)】是周期为( )的( 奇/偶)函数?=-根号2sin2xcos2x=-(根号2/2)sin4x如何来的

问题描述:

函数y=√2 sin(2x-π)cos【2(x +π)】是周期为( )的( 奇/偶)函数?
=-根号2sin2xcos2x
=-(根号2/2)sin4x
如何来的

√2 sin(2x-π)=-√2 sin(π-2x)=-√2sin2x
cos【2(x +π)】=cos(2x+2π)=cos2x
则y=√2 sin(2x-π)cos【2(x +π)】==-√2sin2xcos2x
=-(√2/2)sin4x