Question

42. Sa se calculeze sin $\frac{7 \pi }{6} +cos \frac{2 \pi }{3} .$

Answer 1

[tex]sin \frac{7\pi}{6} =sin \frac{6\pi+\pi}{6}=sin(\pi+ \frac{\pi}{6})=\\ =sin\pi\cdot cos \frac{\pi}{6}+sin \frac{\pi}{6}\cdot cos \pi=\\ =- \frac{1}{2} \\ cos \frac{2\pi}{3} =cos\frac{3\pi-\pi}{3}=cos(\pi-\frac{\pi}{3})=\\ =cos\pi\cdot cos\frac{\pi}{3}+sin\pi\cdot sin \frac{\pi}{3}=\\ =- \frac{1}{2} \\ sin \frac{7\pi}{6} +cos \frac{2\pi}{3}=- \frac{1}{2} - \frac{1}{2} =-1[/tex]