4) cos2 π 12 5) sin2 π 8 Задание 126. Упростите выражение. 1) 1+cos a sin a 2) sin a 1-cos a 3) 1-cos 6a sin 6a 4) sin 4a 1 + cos 4a 5) 1+cos a cos a-1 6) 1-cos 2a 1+cos 2a ctg a

\[\frac{1 + \cos \alpha}{\sin \alpha} = \frac{2 \cos^2(\frac{\alpha}{2})}{2 \sin(\frac{\alpha}{2})\cos(\frac{\alpha}{2})} = \frac{\cos(\frac{\alpha}{2})}{\sin(\frac{\alpha}{2})} = ctg(\frac{\alpha}{2})\]

\[\frac{\sin \alpha}{1 - \cos \alpha} = \frac{2 \sin(\frac{\alpha}{2})\cos(\frac{\alpha}{2})}{2 \sin^2(\frac{\alpha}{2})} = \frac{\cos(\frac{\alpha}{2})}{\sin(\frac{\alpha}{2})} = ctg(\frac{\alpha}{2})\]

\[\frac{1 - \cos 6\alpha}{\sin 6\alpha} = \frac{2 \sin^2(3\alpha)}{2 \sin(3\alpha)\cos(3\alpha)} = \frac{\sin(3\alpha)}{\cos(3\alpha)} = tg(3\alpha)\]

\[\frac{\sin 4\alpha}{1 + \cos 4\alpha} = \frac{2 \sin(2\alpha)\cos(2\alpha)}{2 \cos^2(2\alpha)} = \frac{\sin(2\alpha)}{\cos(2\alpha)} = tg(2\alpha)\]

\[\frac{1 + \cos \alpha}{\cos \alpha - 1} = -\frac{1 + \cos \alpha}{1 - \cos \alpha} = -\frac{2 \cos^2(\frac{\alpha}{2})}{2 \sin^2(\frac{\alpha}{2})} = -ctg^2(\frac{\alpha}{2})\]

\[\frac{1 - \cos 2\alpha}{1 + \cos 2\alpha} \cdot ctg \alpha = \frac{2 \sin^2 \alpha}{2 \cos^2 \alpha} \cdot ctg \alpha = tg^2 \alpha \cdot ctg \alpha = \frac{\sin^2 \alpha}{\cos^2 \alpha} \cdot \frac{\cos \alpha}{\sin \alpha} = \frac{\sin \alpha}{\cos \alpha} = tg \alpha\]