Can you prove sin4x=(4sinxcosx)(2cos^2x-1)? I just can't figure it out, thank you in advance!!!
sin 4x = (4sinxcosx) (2cos^2 (2x) - 1 )
4sinxcosx = 2*2sinxcosx = 2sin2x
2cos^2(2x) - 1 = cos(4x)
But.....as the graph below shows,
sin (4x) is not equal to 2sin(2x)cos(4x)
So....this is not an identity.....
Can you prove sin4x=(4sinxcosx)(2cos^2x-1)?
sin(4x)=[4sin(x)cos(x)][2cos2(x)−1]cos(2x)=cos2(x)−sin2(x)=cos2(x)−[1−cos2(x)]=2cos2(x)−1sin(4x)=[4sin(x)cos(x)][cos(2x)]sin(2x)=2⋅sin(x)cos(x)sin(4x)=2sin(2x)cos(2x)sin(4x)=sin(2x+2x)=sin(2x)cos(2x)+cos(2x)sin(2x)=2sin(2x)cos(2x)sin(4x)=sin(4x) ✓