Упростить выражение: 506 1) 2 cos 40° * cos 50°; 2) 2 sin 25° * sin 65°; 3) sin 2a + (sin a - cos a)^2; 4) cos 4a + sin^2 2a. 507 1) sin 2a / ((sin a + cos a)^2 - 1); 2) (1 + cos 2a) / (1 - cos 2a).

506 1) $2 \cos 40^{\circ} \cdot \cos 50^{\circ} = 2 \cos 40^{\circ} \cdot \sin (90^{\circ} - 50^{\circ}) = 2 \cos 40^{\circ} \cdot \sin 40^{\circ} = \sin (2 \cdot 40^{\circ}) = \sin 80^{\circ}$ 2) $2 \sin 25^{\circ} \cdot \sin 65^{\circ} = 2 \sin 25^{\circ} \cdot \cos (90^{\circ} - 65^{\circ}) = 2 \sin 25^{\circ} \cdot \cos 25^{\circ} = \sin (2 \cdot 25^{\circ}) = \sin 50^{\circ}$ 3) $\sin 2\alpha + (\sin \alpha - \cos \alpha)^2 = \sin 2\alpha + \sin^2 \alpha - 2 \sin \alpha \cos \alpha + \cos^2 \alpha = \sin 2\alpha + 1 - \sin 2\alpha = 1$ 4) $\cos 4\alpha + \sin^2 2\alpha = (1 - 2 \sin^2 2\alpha) + \sin^2 2\alpha = 1 - \sin^2 2\alpha = \cos^2 2\alpha$ 507 1) $\frac{\sin 2\alpha}{(\sin \alpha + \cos \alpha)^2 - 1} = \frac{\sin 2\alpha}{\sin^2 \alpha + 2 \sin \alpha \cos \alpha + \cos^2 \alpha - 1} = \frac{\sin 2\alpha}{1 + \sin 2\alpha - 1} = \frac{\sin 2\alpha}{\sin 2\alpha} = 1$ 2) $\frac{1 + \cos 2\alpha}{1 - \cos 2\alpha} = \frac{2 \cos^2 \alpha}{2 \sin^2 \alpha} = \text{ctg}^2 \alpha$ Ответ: 506: 1) $\sin 80^{\circ}$; 2) $\sin 50^{\circ}$; 3) 1; 4) $\cos^2 2\alpha$. 507: 1) 1; 2) $\text{ctg}^2 \alpha$.