Question

447(4,6) и 448(1)
пожалуйста)))

Answer 1

$\sin \frac{7 \pi }{12} =\sin( \frac{ \pi }{3}+\frac{ \pi }{4}) =\sin\frac{ \pi }{3}\cos\frac{ \pi }{4}+\cos\frac{ \pi }{3}\sin\frac{ \pi }{4}= \\\ =\frac{ \sqrt{3} }{2} \cdot \frac{ \sqrt{2} }{2}+ \frac{1}{2} \cdot \frac{ \sqrt{2} }{2}= \frac{ \sqrt{6}+ \sqrt{2} }{4}$

$ctg15= \frac{1}{tg15} = \frac{1}{tg(45-30)} =\frac{1+tg45tg30}{tg45-tg30} = \\\ =\frac{1+1\cdot \frac{ \sqrt{3} }{3} }{1- \frac{ \sqrt{3} }{3}} = \frac{(1+\frac{ \sqrt{3} }{3} )^2}{(1- \frac{ \sqrt{3} }{3})(1+\frac{ \sqrt{3} }{3})} = \frac{1+\frac{ 2\sqrt{3} }{3} + \frac{1}{3} }{1- \frac{1 }{3}} = \frac{\frac{4}{3}+\frac{ 2\sqrt{3} }{3} }{ \frac{2 }{3}} = \frac{4+ 2\sqrt{3} }{2} =2+ \sqrt{3}$

$\sin \alpha = \sqrt{1-\cos^2 \alpha } =\sqrt{1-( \frac{5}{13} )^2 } = \frac{12}{13} \\\ \cos \beta =- \sqrt{1-\sin^2 \beta } =- \sqrt{1-(-0.6)^2 } =- 0.8 \\\ \sin( \alpha - \beta )=\sin \alpha \cos \beta -\cos \alpha \sin \beta = \\\ = \frac{12}{13} \cdot (- \frac{8}{10} )- \frac{5}{13} \cdot (- \frac{6}{10} )= -\frac{12}{13} \cdot \frac{4}{5} + \frac{5}{13} \cdot \frac{3}{5} = -\frac{48}{65} + \frac{15}{65} = -\frac{33}{65}$