Lượng giác

J

jet_nguyen

Gợi ý:
Câu 1:
sin⁡(2x+π3)=cos⁡2x\sin (2x+\dfrac{\pi}{3})=\cos 2x⟺sin⁡(2x+π3)=sin⁡(π2−2x)\Longleftrightarrow \sin (2x+\dfrac{\pi}{3}) =\sin (\dfrac{\pi}{2}-2x)
Câu 2:
sin⁡2x=cos⁡(x−2π3)\sin 2x=\cos (x-\dfrac{2\pi}{3})⟺cos⁡(π2−2x)=cos⁡(x−2π3)\Longleftrightarrow \cos(\dfrac{\pi}{2}-2x)=\cos (x-\dfrac{2\pi}{3})
Tới đây thì cơ bản rồi nhé. ;)
 
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