Giải Bài tập 1.31 trang 41 sgk Toán 11 tập 1 Kết nối

Vì $\frac{\pi }{2}<\alpha <\pi $ nên sin α > 0. Mặt khác từ $sin^{2} α + cos^{2} α = 1$ suy ra $sin\alpha =\sqrt{1-có^{2}\alpha }=\sqrt{1-(-\frac{1}{\sqrt{3}})^{2}}=\frac{\sqrt{6}}{3}$

a) $sin(\alpha +\frac{\pi }{6})=sin\alpha cos\frac{\pi }{6}+cos\alpha sin\frac{\pi }{6}=\frac{\sqrt{6}}{3}\times \frac{\sqrt{3}}{2}+(-\frac{1}{\sqrt{3}})\times \frac{1}{2}=\frac{3\sqrt{2}-\sqrt{3}}{6}$

b) $cos(\alpha +\frac{\pi }{6})=cos\alpha cos\frac{\pi }{6}-sin\alpha sin\frac{\pi }{6}=-\frac{1}{\sqrt{3}}\times \frac{\sqrt{3}}{2}-\frac{\sqrt{6}}{3}\times \frac{1}{2}=\frac{-3-\sqrt{6}}{6}$

c) $sin(\alpha -\frac{\pi }{3})=sin\alpha cos\frac{\pi }{3}-cos\alpha sin\frac{\pi }{3}=\frac{\sqrt{6}}{3}\times \frac{1}{2}-(-\frac{1}{\sqrt{3}})\times \frac{\sqrt{3}}{2}=\frac{\sqrt{6}+3}{6}$

d) $cos(\alpha -\frac{\pi }{6})=cos\alpha cos\frac{\pi }{6}+sin\alpha sin\frac{\pi }{6}=-\frac{1}{\sqrt{3}}\times \frac{\sqrt{3}}{2}+\frac{\sqrt{6}}{3}\times \frac{1}{2}=\frac{-3+\sqrt{6}}{6}$