Question

CMR

\(4sinx.sin\left(\dfrac{\pi}{3}+x\right).sin\left(\dfrac{\pi}{3}-x\right)=sin3x ;\forall x\in R\)

Answer 1

Lời giải:

Sử dụng công thức lượng giác:

\(\cos a-\cos b=(-2)\sin \frac{a+b}{2}\sin \frac{a-b}{2}\) ta có:

\(\cos \frac{2\pi}{3}-\cos 2x=-2\sin \left(\frac{\pi}{3}+x\right)\sin \left(\frac{\pi}{3}-x \right)\)

Suy ra:

\(\sin \left(\frac{\pi}{3}+x\right)\sin \left(\frac{\pi}{3}-x \right)=\frac{\cos \frac{2\pi}{3}-\cos 2x}{-2}=\frac{1+2\cos 2x}{4}\)

\(\Rightarrow \text{VT}=4\sin x\sin \left(\frac{\pi}{3}+x\right)\sin \left(\frac{\pi}{3}-x \right)=\sin x(1+2\cos 2x)\)

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

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

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

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

Do đó ta có đpcm.