Question

Chứng minh các đẳng thứ sau:

\(1,sin^8x-cos^8x=-(\dfrac{7}{8}cos2x+\dfrac{1}{8}cos6x) \)

2\(sin^2x×cos^4x=\dfrac{1}{16}+\dfrac{1}{32}cos2x-\dfrac{1}{16}cos4x-\dfrac{1}{32}cos6x\)

Answer 1

1.

\(\sin ^8x-\cos ^8x=(\sin ^4x+\cos ^4x)(\sin ^4x-\cos ^4x)\)

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

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

\(=(1-\frac{\sin ^22x}{2})(-\cos 2x)=-\frac{(2-\sin ^22x)\cos 2x}{2}=-\frac{(1+\cos ^22x)\cos 2x}{2}\) (1)

\(-(\frac{7}{8}\cos 2x+\frac{1}{8}\cos 6x)=\frac{-7}{8}\cos 2x-\frac{1}{8}(4\cos ^32x-3\cos 2x)=-\frac{\cos 2x+\cos ^32x}{2}\)

\(=\frac{-\cos 2x(\cos ^22x+1)}{2}\) (2)

Từ $(1);(2)$ ta có đpcm.

Answer 2

2.

\(\text{VP}=\frac{1}{32}(2+\cos 2x-2\cos 4x-\cos 6x)\)

\(=\frac{1}{32}[2+\cos 2x-2(2\cos ^22x-1)-(4\cos ^32x-3\cos 2x)]\)

\(=\frac{1}{8}(-\cos ^32x-\cos ^22x+\cos 2x+1)=\frac{1}{8}(\cos 2x+1)(1-\cos ^22x)=\frac{1}{8}(\cos 2x+1)\sin ^22x\) (1)

\(\text{VT}=\sin ^2x\cos ^4x=\frac{1}{8}.(2\sin x\cos x)^2.2\cos ^2x=\frac{1}{8}\sin ^22x.(\cos 2x+1)(2)\)

Từ $(1);(2)$ ta có đpcm.