Ответ:
- \(2\cos20^\circ\cos40^\circ-\cos20^\circ=\cos60^\circ+\cos20^\circ-\cos20^\circ=\frac12\).
- \(\sin\alpha(1+2\cos2\alpha)=\sin\alpha+(\sin3\alpha-\sin\alpha)=\sin3\alpha\), так как \(2\sin\alpha\cos2\alpha=\sin3\alpha-\sin\alpha\).
- \(2\cos\alpha\cos2\alpha-\cos3\alpha=(\cos3\alpha+\cos\alpha)-\cos3\alpha=\cos\alpha\).
- \(\cos2\alpha+2\sin(\alpha+30^\circ)\sin(\alpha-30^\circ)=\cos2\alpha+\cos60^\circ-\cos2\alpha=\frac12\).
Ответ: 1) \(\frac12\); 2) \(\sin3\alpha\); 3) \(\cos\alpha\); 4) \(\frac12\).
