Hãy xác định hệ thức sai: sinx.cos^3 x - cosx.sin^3 x = sin4x/4

* Hướng dẫn giải

Đáp án C

$$\begin{gathered} +)\sin \text{x}{\cos }^{3}x-\cos x{\sin }^{3}x=\mathrm{s}\text{inx}\frac{3\cos x+\cos 3x}{4}-\cos x.\frac{3\sin x-\sin 3x}{4} \\ =\frac{3}{4}\sin \text{x}\cos x+\frac{1}{4}\sin x\cos 3x-\frac{3}{4}\sin \text{x}\cos x+\frac{1}{4}\sin 3x\cos x \\ =\frac{1}{4}\left(\sin x\cos 3x+\sin 3x\cos x\right) \\ =\frac{1}{4}\sin \left(x+3x\right)=\frac{\sin 4x}{4} \end{gathered}$$

$$\begin{gathered} +){\sin }^{4}x+{\cos }^{4}x={\left({\sin }^{2}x+{\cos }^{2}x\right)}^2-2{\sin }^{2}x{\cos }^{2}x=1-\frac{1}{2}{\sin }^{2}2x \\ =1-\frac{1}{2}-\frac{1-\cos 4x}{2}=\frac{3+\cos 4x}{4} \end{gathered}$$

$$\begin{gathered} +){\cot }^{2}x+{\tan }^{2}x=\frac{{\cos }^{2}x}{{\sin }^{2}x}+\frac{{\sin }^{2}x}{{\cos }^{2}x}=\frac{{\cos }^{4}x+{\sin }^{4}x}{{\sin }^{2}x{\cos }^{2}x} \\ =\frac{\frac{3+\cos 4x}{4}}{\frac{1}{4}{\sin }^{2}2x}=\frac{3+\cos 4x}{\frac{1}{2}\left(1-\cos 4x\right)}=\frac{2\cos 4x+6}{1-\cos 4x} \end{gathered}$$