[Toán 10] Tính

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ngocthao1995

Đặt [TEX]x=\frac{\pi}{5}[/TEX]

[TEX]\Rightarrow 2x+3x=\pi[/TEX]

Ta có

[TEX]sin2x=sin3x \\ \Leftrightarrow 2sinxcosx=3sinx-4sin^3x\\ \Leftrightarrow 2cosx=3-4+4cos^2x \\ \Leftrightarrow 4cos^2x-2cosx-1=0\\ \Leftrightarrow cosx=\frac{1-\sqrt{5}}{4} \ \text{hoac}\ cosx=\frac{1+\sqrt{5}}{4}\\ \Rightarrow cos{\frac{\pi}{5}}-cos{\frac{2\pi}{5}}=\frac{1+\sqrt{5}}{4}-\frac{\sqrt{5}-1}{4}=\frac{1}{2}[/TEX]
 
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