Quay lại đáp án

b. Ta có: $\sin 54^0=\cos 36^0 \Leftrightarrow \sin3.18^0=\cos 2.18^0$$\Leftrightarrow 3.\sin 18^0-4\sin^3 18^0=2\sin^2 18^0-1$$\Leftrightarrow 4\sin^3 18^0-2\sin^2 18^0-3\sin 18^0+1=0 $$\Leftrightarrow (\sin 18^0-1)(4\sin^2 18^0+2\sin 18^0-1)=0 (1)$Do $0<\sin 18^0<1$ nên $4\sin^2 18^0+2\sin 18^0-1=0\Leftrightarrow \sin18^o=\frac{-1+\sqrt5}{4}$Và do đó suy ra $\cos 18^0=\sqrt{1-\sin^2 18^0}=\sqrt{\frac{5}{8}+\frac{\sqrt5}{8}}$

b. Ta có: $\sin 54^0=\cos 36^0 \Leftrightarrow \sin3.18^0=\cos 2.18^0$$\Leftrightarrow 3.\sin 18^0-4\sin^3 18^0=1-2\sin^2 18^0$$\Leftrightarrow 4\sin^3 18^0-2\sin^2 18^0-3\sin 18^0+1=0 $$\Leftrightarrow (\sin 18^0-1)(4\sin^2 18^0+2\sin 18^0-1)=0 (1)$Do $0<\sin 18^0<1$ nên $4\sin^2 18^0+2\sin 18^0-1=0\Leftrightarrow \sin18^o=\frac{-1+\sqrt5}{4}$Và do đó suy ra $\cos 18^0=\sqrt{1-\sin^2 18^0}=\sqrt{\frac{5}{8}+\frac{\sqrt5}{8}}$

b. Ta có: $\sin 54^0=\cos 36^0 \Leftrightarrow \sin3.18^0=\cos 2.18^0$$\Leftrightarrow 3.\sin 18^0-4\sin^3 18^0=2\sin^2 18^0-1$$\Leftrightarrow 4\sin^3 18^0-2\sin^2 18^0-3\sin 18^0+1=0 $$\Leftrightarrow (\sin 18^0-1)(4\sin^2 18^0+2\sin 18^0-1)=0 (1)$Do $0<\sin 18^0<1$ nên $\cos 18^0=\sqrt{1-\sin^2 18^0}=\frac{\sqrt{10+2\sqrt{5} } }{4}$Và do đó suy ra $\cos 18^0=\sqrt{1-\sin^2 18^0}=\frac{10+2 \sqrt{5} }{4}$

b. Ta có: $\sin 54^0=\cos 36^0 \Leftrightarrow \sin3.18^0=\cos 2.18^0$$\Leftrightarrow 3.\sin 18^0-4\sin^3 18^0=2\sin^2 18^0-1$$\Leftrightarrow 4\sin^3 18^0-2\sin^2 18^0-3\sin 18^0+1=0 $$\Leftrightarrow (\sin 18^0-1)(4\sin^2 18^0+2\sin 18^0-1)=0 (1)$Do $0<\sin 18^0<1$ nên $4\sin^2 18^0+2\sin 18^0-1=0\Leftrightarrow \sin18^o=\frac{-1+\sqrt5}{4}$Và do đó suy ra $\cos 18^0=\sqrt{1-\sin^2 18^0}=\sqrt{\frac{5}{8}+\frac{\sqrt5}{8}}$