Решение тригонометрических выражений с sin, cos, tg, ctg

Задание 4. Вычислите: 1) \( 2\sin 30^\circ - \text{tg } 45^\circ + \text{ctg } 30^\circ = 2 \cdot \frac{1}{2} - 1 + \sqrt{3} = 1 - 1 + \sqrt{3} = \sqrt{3} \) 2) \( \text{tg } 60^\circ + 2\cos 45^\circ - \sqrt{3} \text{ctg } 45^\circ = \sqrt{3} + 2 \cdot \frac{\sqrt{2}}{2} - \sqrt{3} \cdot 1 = \sqrt{3} + \sqrt{2} - \sqrt{3} = \sqrt{2} \) 3) \( 6\cos 30^\circ - 3\text{tg } 60^\circ + 2\sin 45^\circ = 6 \cdot \frac{\sqrt{3}}{2} - 3 \cdot \sqrt{3} + 2 \cdot \frac{\sqrt{2}}{2} = 3\sqrt{3} - 3\sqrt{3} + \sqrt{2} = \sqrt{2} \) 4) \( \sqrt{3} \text{tg } 30^\circ + 4\sin 30^\circ - \sqrt{3} \text{ctg } 30^\circ = \sqrt{3} \cdot \frac{\sqrt{3}}{3} + 4 \cdot \frac{1}{2} - \sqrt{3} \cdot \sqrt{3} = \frac{3}{3} + 2 - 3 = 1 + 2 - 3 = 0 \) 5) \( \sqrt{3} \sin \frac{\pi}{3} - 2\cos \frac{\pi}{6} + \frac{\sqrt{3}}{2} \text{tg } \frac{\pi}{3} = \sqrt{3} \cdot \frac{\sqrt{3}}{2} - 2 \cdot \frac{\sqrt{3}}{2} + \frac{\sqrt{3}}{2} \cdot \sqrt{3} = \frac{3}{2} - \sqrt{3} + \frac{3}{2} = 3 - \sqrt{3} \) 6) \( 2\cos \frac{\pi}{3} + 2\sin \frac{\pi}{6} - 2\sin \frac{\pi}{4} = 2 \cdot \frac{1}{2} + 2 \cdot \frac{1}{2} - 2 \cdot \frac{\sqrt{2}}{2} = 1 + 1 - \sqrt{2} = 2 - \sqrt{2} \) 7) \( \sqrt{3} \cos \frac{\pi}{6} + 2\sin \frac{\pi}{3} - \frac{\sqrt{3}}{2} \text{ctg } \frac{\pi}{6} = \sqrt{3} \cdot \frac{\sqrt{3}}{2} + 2 \cdot \frac{\sqrt{3}}{2} - \frac{\sqrt{3}}{2} \cdot \sqrt{3} = \frac{3}{2} + \sqrt{3} - \frac{3}{2} = \sqrt{3} \) 8) \( \sqrt{2} \cos \frac{\pi}{4} - 2\sin \frac{\pi}{6} + \text{ctg } \frac{\pi}{6} = \sqrt{2} \cdot \frac{\sqrt{2}}{2} - 2 \cdot \frac{1}{2} + \sqrt{3} = \frac{2}{2} - 1 + \sqrt{3} = 1 - 1 + \sqrt{3} = \sqrt{3} \) 9) \( 2\sin \pi - \cos 0 + \text{tg } 0 + 3\cos \frac{\pi}{2} - \sin \frac{3\pi}{2} = 2 \cdot 0 - 1 + 0 + 3 \cdot 0 - (-1) = 0 - 1 + 0 + 0 + 1 = 0 \) 10) \( 5\sin 90^\circ + 2\cos 0^\circ - 2\sin 270^\circ + 10\cos 180^\circ = 5 \cdot 1 + 2 \cdot 1 - 2 \cdot (-1) + 10 \cdot (-1) = 5 + 2 + 2 - 10 = 9 - 10 = -1 \)