Вычисления и преобразования

Вынесем минус из числителя чтобы получить косинус двойного угла в скобках:

$$\frac{-24 \cos 34^\circ}{\cos 34^\circ} = -24$$

$$\sin 61^\circ = \sin(90^\circ - 29^\circ) = \cos 29^\circ$$ $$\frac{5 \cos 29^\circ}{\cos 29^\circ} = 5$$

$$\cos \frac{\pi}{4} = \frac{\sqrt{2}}{2}, \quad \cos \frac{7\pi}{3} = \cos \left(2\pi + \frac{\pi}{3}\right) = \cos \frac{\pi}{3} = \frac{1}{2}$$ $$4\sqrt{2} \cdot \frac{\sqrt{2}}{2} \cdot \frac{1}{2} = 4\sqrt{2} \cdot \frac{\sqrt{2}}{4} = \sqrt{2} \cdot \sqrt{2} = 2$$

$$tg \frac{\pi}{6} = \frac{1}{\sqrt{3}}, \quad \sin \frac{\pi}{4} = \frac{\sqrt{2}}{2}$$ $$36\sqrt{6} \cdot \frac{1}{\sqrt{3}} \cdot \frac{\sqrt{2}}{2} = 36\sqrt{6} \cdot \frac{\sqrt{2}}{2\sqrt{3}} = 18 \cdot \frac{\sqrt{12}}{\sqrt{3}} = 18 \cdot \frac{2\sqrt{3}}{\sqrt{3}} = 36$$

$$\sin\left(-\frac{27\pi}{4}\right) = -\sin\left(\frac{27\pi}{4}\right) = -\sin\left(6\pi + \frac{3\pi}{4}\right) = -\sin\left(\frac{3\pi}{4}\right) = -\frac{\sqrt{2}}{2}$$ $$\cos\left(\frac{31\pi}{4}\right) = \cos\left(8\pi - \frac{\pi}{4}\right) = \cos\left(-\frac{\pi}{4}\right) = \cos\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}$$ $$\frac{8}{\left(-\frac{\sqrt{2}}{2}\right)\cdot\left(\frac{\sqrt{2}}{2}\right)} = \frac{8}{-\frac{2}{4}} = \frac{8}{-\frac{1}{2}} = -16$$