Question

\(Cos3x.SIN^3X+SIN3X.COS^3X=\frac{3}{4}.SIN4X\)

Answer 1

ta có:( theo công thức lượng giác nhân ba)

VT= cos3x.sin³x+sin3x.cos³x=cos3x.\(\frac{3sinx-sin3x}{4}\)^{+sin3x\(\frac{3cosx+cos3x}{4}\)}

=\(\frac{3}{4}\)((sinx-\(\frac{1}{3}\)sin3x).cos3x+sin3x(cosx+\(\frac{1}{3}\)cos3x))

^{.=\(\frac{3}{4}\)}(cos3x.sinx-\(\frac{1}{3}\)sin3x.cos3x+sin3x.cosx+\(\frac{1}{3}\)sin3x.cos3x)

=\(\frac{3}{4}\)(sinx.cos3x+cosx.sin3x)

=\(\frac{3}{4}\)sin(x+3x)=\(\frac{3}{4}\)sin4x

=> đpcm