Ta có
$\sqrt 2 -\sin x -\cos x = \sqrt 2 - \sqrt 2 \sin (x+\dfrac{\pi}{4})=\sqrt 2 \bigg [1-\sin (x+\dfrac{\pi}{4}) \bigg ]$
$\sin x -\cos x = \sqrt 2 \sin (x-\dfrac{\pi}{4})=-\sqrt 2 \cos (x+\dfrac{\pi}{4})$
Bạn chọn cái nào thích thì lắp vào
Câu 1
$\dfrac{\dfrac{\sqrt 3}{2} -\sin 3x}{\dfrac{\sqrt 3}{2} +\sin 3x}=\dfrac{\sin \dfrac{\pi}{3} -\sin 3x}{\sin \dfrac{\pi}{3} +\sin 3x}$
$=\dfrac{2\cos \dfrac{3x+\dfrac{\pi}{3}}{2} \sin \dfrac{\dfrac{\pi}{3}-3x}{2}}{2\sin \dfrac{3x+\dfrac{\pi}{3}}{2} \cos \dfrac{\dfrac{\pi}{3}-3x}{2}}=\cot \dfrac{3x+\dfrac{\pi}{3}}{2} . \tan \dfrac{\dfrac{\pi}{3}-3x}{2}$