Решение: Применение формул двойного угла

Практическая работа: Применение формул двойного угла. Задание 1. Упростите выражения: 1) \(\frac{\sin 2t}{\cos t} - \sin t = \frac{2 \sin t \cos t}{\cos t} - \sin t = 2 \sin t - \sin t = \sin t\) 2) \(\frac{\sin 6t}{\cos^2 3t} = \frac{2 \sin 3t \cos 3t}{\cos^2 3t} = \frac{2 \sin 3t}{\cos 3t} = 2 \text{tg } 3t\) 3) \(\cos^2 t - \cos 2t = \cos^2 t - (\cos^2 t - \sin^2 t) = \cos^2 t - \cos^2 t + \sin^2 t = \sin^2 t\) 4) \(\frac{\cos 2t}{\cos t - \sin t} - \sin t = \frac{(\cos t - \sin t)(\cos t + \sin t)}{\cos t - \sin t} - \sin t = \cos t + \sin t - \sin t = \cos t\) 5) \(\frac{\sin 40^\circ}{\sin 20^\circ} = \frac{2 \sin 20^\circ \cos 20^\circ}{\sin 20^\circ} = 2 \cos 20^\circ\) 6) \(\frac{\sin 100^\circ}{2 \cos 50^\circ} = \frac{2 \sin 50^\circ \cos 50^\circ}{2 \cos 50^\circ} = \sin 50^\circ\) 7) \(\frac{\cos 80^\circ}{\cos 40^\circ + \sin 40^\circ} = \frac{\cos^2 40^\circ - \sin^2 40^\circ}{\cos 40^\circ + \sin 40^\circ} = \frac{(\cos 40^\circ - \sin 40^\circ)(\cos 40^\circ + \sin 40^\circ)}{\cos 40^\circ + \sin 40^\circ} = \cos 40^\circ - \sin 40^\circ\) 8) \(\frac{\cos 36^\circ + \sin^2 18^\circ}{\cos 18^\circ} = \frac{(1 - 2 \sin^2 18^\circ) + \sin^2 18^\circ}{\cos 18^\circ} = \frac{1 - \sin^2 18^\circ}{\cos 18^\circ} = \frac{\cos^2 18^\circ}{\cos 18^\circ} = \cos 18^\circ\) 9) \(\frac{\sin t}{2 \cos^2(\frac{t}{2})} = \frac{2 \sin(\frac{t}{2}) \cos(\frac{t}{2})}{2 \cos^2(\frac{t}{2})} = \frac{\sin(\frac{t}{2})}{\cos(\frac{t}{2})} = \text{tg}(\frac{t}{2})\) 10) \(\frac{\sin 4t}{\cos 2t} = \frac{2 \sin 2t \cos 2t}{\cos 2t} = 2 \sin 2t\) Задание 2. Найдите значения выражения: 1) \(2 \sin 15^\circ \cos 15^\circ = \sin(2 \cdot 15^\circ) = \sin 30^\circ = \frac{1}{2}\) 2) \((\cos 75^\circ - \sin 75^\circ)^2 = \cos^2 75^\circ - 2 \sin 75^\circ \cos 75^\circ + \sin^2 75^\circ = 1 - \sin 150^\circ = 1 - \frac{1}{2} = \frac{1}{2}\) 3) \(\cos^2 15^\circ - \sin^2 15^\circ = \cos(2 \cdot 15^\circ) = \cos 30^\circ = \frac{\sqrt{3}}{2}\) 4) \((\cos 15^\circ - \sin 15^\circ)^2 = \cos^2 15^\circ - 2 \sin 15^\circ \cos 15^\circ + \sin^2 15^\circ = 1 - \sin 30^\circ = 1 - \frac{1}{2} = \frac{1}{2}\) 5) \(2 \sin \frac{\pi}{8} \cos \frac{\pi}{8} = \sin(2 \cdot \frac{\pi}{8}) = \sin \frac{\pi}{4} = \frac{\sqrt{2}}{2}\) 6) \(\sin \frac{\pi}{8} \cos \frac{\pi}{8} + \frac{1}{4} = \frac{1}{2} \sin \frac{\pi}{4} + \frac{1}{4} = \frac{1}{2} \cdot \frac{\sqrt{2}}{2} + \frac{1}{4} = \frac{\sqrt{2} + 1}{4}\) 7) \(\cos^2 \frac{\pi}{8} - \sin^2 \frac{\pi}{8} = \cos(2 \cdot \frac{\pi}{8}) = \cos \frac{\pi}{4} = \frac{\sqrt{2}}{2}\) 8) \(\frac{\sqrt{2}}{2} - (\cos \frac{\pi}{8} + \sin \frac{\pi}{8})^2 = \frac{\sqrt{2}}{2} - (1 + \sin \frac{\pi}{4}) = \frac{\sqrt{2}}{2} - (1 + \frac{\sqrt{2}}{2}) = -1\) 9) \(\frac{2 \text{tg } \frac{\pi}{8}}{1 - \text{tg}^2 \frac{\pi}{8}} = \text{tg}(2 \cdot \frac{\pi}{8}) = \text{tg } \frac{\pi}{4} = 1\) 10) \(\frac{2 \text{tg } 75^\circ}{1 - \text{tg}^2 75^\circ} = \text{tg}(2 \cdot 75^\circ) = \text{tg } 150^\circ = -\frac{\sqrt{3}}{3}\)