Đặt \(sin\frac{\pi}{10}=x\)
\(\Rightarrow cos2x=cos\left(\frac{2\pi}{10}\right)=sin\left(\frac{\pi}{2}-\frac{2\pi}{10}\right)=sin\left(\frac{3\pi}{10}\right)=sin3x\Rightarrow\frac{cos2x}{sin3x}=1\)
Ta có:
\(\frac{1}{sinx}-\frac{1}{sin3x}=\frac{sin3x-sinx}{sinx.sin3x}=\frac{2cos2x.sinx}{sinx.sin3x}=\frac{2cos2x}{sin3x}=2.1=2\)