1. Prove that \( 2 \sin 5 \alpha \cos 4 \alpha-\sin 9 \alpha=\sin \alpha \). 2. If \( \sin x=\frac{2}{5} \), find the value of \( \cos 2 x \) without using a calculator. 3. If \( \cos 2 x=0,28 \), find the value of \( \cos x \) without using a cal culator.

1. We start with the identity \[ 2\sin A \cos B = \sin (A+B) + \sin (A-B). \] Let \( A=5\alpha \) and \( B=4\alpha \). Then \[ 2\sin5\alpha \cos4\alpha = \sin(5\alpha+4\alpha)+\sin(5\alpha-4\alpha) = \sin 9\alpha + \sin \alpha. \] Subtract \(\sin9\alpha\) from both sides to obtain \[ 2\sin5\alpha \cos4\alpha - \sin9\alpha = \sin \alpha. \] 2. Given \(\sin x=\frac{2}{5}\), we use the double-angle formula for cosine: \[ \cos2x = 1-2\sin^2x. \] First, compute \[ \sin^2x=\left(\frac{2}{5}\right)^2 = \frac{4}{25}. \] Then, substitute into the formula: \[ \cos2x=1-2\cdot\frac{4}{25}=1-\frac{8}{25}=\frac{25}{25}-\frac{8}{25}=\frac{17}{25}. \] 3. Given \(\cos2x=0.28\), write \(0.28\) as a fraction: \[ 0.28=\frac{28}{100}=\frac{7}{25}. \] Recall the cosine double angle formula: \[ \cos2x=2\cos^2x-1. \] Set up the equation: \[ 2\cos^2x-1=\frac{7}{25}. \] Solve for \(\cos^2x\): \[ 2\cos^2x=\frac{7}{25}+1=\frac{7}{25}+\frac{25}{25}=\frac{32}{25}\quad\Longrightarrow\quad \cos^2x=\frac{32}{25}\cdot\frac{1}{2}=\frac{16}{25}. \] Taking the square root gives: \[ \cos x=\pm\sqrt{\frac{16}{25}}=\pm\frac{4}{5}. \] Thus, the final answers are: 1. \(2\sin5\alpha\cos4\alpha-\sin9\alpha=\sin\alpha\). 2. \(\cos2x=\frac{17}{25}\). 3. \(\cos x=\pm\frac{4}{5}\).