Ta có:
`VT=sinx(1+2cos2x+2cos4x+2cos6x)`
`=sinx+2sinxcos2x+2sinxcos4x+2sinxcos6x`
`=sinx+2. 1/ 2 .[sin(x+2x)+sin(x-2x)]+2. 1/ 2 [sin(x+4x)+sin(x-4x)]+2. 1/ 2 .[sin(x+6x)+sin(x-6x)]`
`=sinx+sin3x-sinx+sin5x-sin3x+sin7x-sin5x`
`=sin7x=VP`
Vậy `sinx(1+2cos2x+2cos4x+2cos6x)=sin7x`