chứng minh đẳng thức sau: $\frac{2cos2x – sin4x}{2cos2x + sin4x}$ =$tan^{2}$( $\frac{ π}{4}$ -x)

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1. Giải thích các bước giải:

   Ta có:

   $\dfrac{2\cos2x – \sin4x}{2\cos2x + \sin4x}$

   $=\dfrac{2\cos2x – 2\sin2x\cos2x}{2\cos2x + 2\sin2x\cos2x}$

   $=\dfrac{1 -\sin2x}{1 + \sin2x}$

   $=\dfrac{\sin^2x+\cos^2x -2\sin x\cos x}{\sin^2x+\cos^2x +2\sin x\cos x}$

   $=\dfrac{\left(\sin x-\cos x\right)^2}{\left(\sin x+\cos x \right)^2}$

   $=\dfrac{\left(\dfrac{\sqrt{2}}{2}\sin x-\dfrac{\sqrt{2}}{2}\cos x\right)^2}{\left(\dfrac{\sqrt{2}}{2}\sin x+\dfrac{\sqrt{2}}{2}\cos x \right)^2}$

   $=\dfrac{\left(\cos\left(\dfrac{\pi}{4}\right)\sin x-\sin\left(\dfrac{\pi}{4}\right)\cos x\right)^2}{\left(\sin\left(\dfrac{\pi}{4}\right)\sin x+\cos x\cos\left(\dfrac{\pi}{4}\right) \right)^2}$

   $=\dfrac{\left(\sin x\cos\left(\dfrac{\pi}{4}\right)-\cos x\sin\left(\dfrac{\pi}{4}\right)\right)^2}{\left(\cos\left(x\right)\cos\left(\dfrac{\pi}{4}\right)+\sin x\sin\left(\dfrac{\pi}{4}\right) \right)^2}$

   $=\dfrac{\sin^2\left(x-\dfrac{\pi}{4}\right)}{\cos^2\left(x-\dfrac{\pi}{4}\right)}$

   $=\tan^2\left(x-\dfrac{\pi}{4}\right)$

   Trả lời