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

\(\frac{tan^3x}{sin^2x}-\frac{1}{sinx.cosx}+\frac{cot^3x}{cos^2x}=tan^3x\left(1+cot^2x\right)-\frac{1}{sinx.cosx}+cot^3x\left(1+tan^2x\right)\)

\(=tan^3x+tanx+cot^3x+cotx-\frac{1}{sinx.cosx}\)

\(=tan^3x+cot^3x+\frac{sinx}{cosx}+\frac{cosx}{sinx}-\frac{1}{sinx.cosx}\)

\(=tan^3x+cot^3x+\frac{sin^2x+cos^2x}{sinx.cosx}-\frac{1}{sinx.cosx}\)

\(=tan^3x+cot^3x\)

với (cosx khác 0)

VT: \(\dfrac{cosx+sinx}{cosx^3}=\dfrac{\dfrac{cosx}{cosx}+\dfrac{sinx}{cosx}}{\dfrac{cosx^3}{cosx}}=\dfrac{1+tanx}{cosx^2}\)

VP:

\(tanx^3+tanx^2+tanx+1=\left(tanx+1\right)\left(tanx^2+1\right)\\ =\left(tanx+1\right).\dfrac{1}{cosx^2+1}\)

Vậy VT=VP

Lời giải:
\(3(\sin x+\cos x)-(\sin x+\cos x)^3=(\sin x+\cos x)[3-(\sin x+\cos x)^2]\)

\(=(\sin x+\cos x)[3-(\sin ^2x+\cos ^2x)-2\sin x\cos x]\)

\(=(\sin x+\cos x)(3-1-2\sin x\cos x)=2(\sin x+\cos x)(1-\sin x\cos x)=2(\sin x+\cos x)(\sin ^2x+\cos ^2x-\sin x\cos x)\)

\(=2(\sin ^3+\cos ^3x)\)

\(\Rightarrow \frac{3(\sin x+\cos x)-(\sin x+\cos x)^3}{2}=\sin ^3x+\cos ^3x\)(đpcm)

\(VT=tan^2x\left(tanx+1\right)+tanx+1=\left(tan^2x+1\right)\left(tanx+1\right)\)

\(=\left(\frac{sin^2x}{cos^2x}+1\right)\left(\frac{sinx}{cosx}+1\right)=\frac{1}{cos^2x}\left(\frac{sinx+cosx}{cosx}\right)=\frac{sinx+cosx}{cos^3x}\)

\(sinx.cos^3x-sin^3x.cosx\)

\(=sinx.cosx\left(cos^2x-sin^2x\right)\)

\(=\dfrac{1}{2}sin2x\left(cos^2x-sin^2x\right)\)

\(=\dfrac{1}{2}sin2x.cos2x\)

\(=\dfrac{sin4x}{4}\)

Lời giải:

\((1+\sin x)(\cot x-\cos x)=(1+\sin x)(\frac{\cos x}{\sin x}-\cos x)=\cos x(1+\sin x).\frac{1-\sin x}{\sin x}\)

\(=\frac{\cos x(1-\sin ^2x)}{\sin x}=\frac{\cos x.\cos ^2x}{\sin x}=\frac{\cos ^3x}{\sin x}\)

\(\left(1+sinx\right)\left(cotx-cosx\right)=\left(1+sinx\right)\left(\dfrac{cosx}{sinx}-cosx\right)\)

\(=cosx\left(1+sinx\right)\left(\dfrac{1-sinx}{sinx}\right)=\dfrac{cosx\left(1-sin^2x\right)}{sinx}=\dfrac{cos^3x}{sinx}\)

Đề bài ko chính xác

\(sin3x=3sinx-4sin^3x\Rightarrow sin^3x=\frac{3sinx-sin3x}{4}\)

\(cos3x=4cos^3x-3cosx\Rightarrow cos^3x=\frac{cos3x+3cosx}{4}\)

\(\Rightarrow sin3x.sin^3x+cos3x.cos^3x=sin3x\left(\frac{3sinx-sin3x}{4}\right)+cos3x\left(\frac{cos3x+3cosx}{4}\right)\)

\(=\frac{3}{4}\left(cos3x.cosx+sin3x.sinx\right)+\frac{1}{4}\left(cos^23x-sin^23x\right)\)

\(=\frac{3}{4}cos2x+\frac{1}{4}cos6x\)

\(=\frac{3}{4}cos2x+\frac{1}{4}\left(4cos^32x-3cos2x\right)\)

\(=cos^32x\)