`a)` Ta có:
`\qquad sin(x+y).sin(x-y)`
`=(sinxcosy+cosxsiny).(sinxcosy-sinycosx)`
`=(sinxcosy)^2-(cosxsiny)^2`
`=sin^2xcos^2y-cos^2xsin^2y` (*)
`=sin^2xcos^2y-(1-sin^2x).sin^2y`
`=sin^2x cos^2y+sin^2xsin^2y-sin^2y`
`=sin^2x (cos^2y+sin^2y)-sin^2y`
`=sin^2x.1-sin^2y`
`=sin^2x-sin^2y`
`=(1-cos^2x)-(1-cos^2y)`
`=1-cos^2x-1+cos^2y`
`=cos^2y-cos^2x`
Vậy: `sin(x+y).sin(x-y)`
`=sin^2x-sin^2y`
`=cos^2y-cos^2x`
$\\$
`b)` Ta có:
`\qquad cos(x+y).cos(x-y)`
`=(cosxcosy-sinxsiny).(cosxcosy+sinxsiny)`
`=(cosxcosy)^2-(sinxsiny)^2`
`=cos^2x cos^2y-sin^2xsin^2y`
`=cos^2x cos^2y-(1-cos^2x)sin^2y`
`=cos^2xcos^2y+cos^2xsin^2y-sin^2y`
`=cos^2x(cos^2y+sin^2y)-sin^2y`
`=cos^2x.1-sin^2y`
`=cos^2x-sin^2y`
`=(1-sin^2x)-(1-cos^2y)`
`=1-sin^2x-1+cos^2y`
`=cos^2y-sin^2x`
Vậy: `cos(x+y)cos(x-y)`
`=cos^2x-sin^2y`
`=cos^2y-sin^2x`
