Question

помогите пожалуйста решить очень нужно

Answer 1

Объяснение:

1.

$cos(\alpha +\frac{\pi }{3})\ \ \ \ cos\alpha =\frac{15}{17} \ \ \ \ \ \frac{3\pi }{2} < \alpha < 2\pi \\ sin^2\alpha +cos^2\alpha =1\\sin^2\alpha =1-cos^2\alpha =1-(\frac{15}{17})^2=1-\frac{225}{289}=\frac{289-225}{289}=\frac{64}{289}. \ \ \ \ \\sin\alpha =б\sqrt{\frac{25}{289} } =б\frac{5}{17}. \ \ \ \ \frac{3\pi }{2} < \alpha < 2\pi \ \ \ \ \Rightarrow\\sin\alpha =-\frac{5}{17}.$

$cos(\alpha +\frac{\pi }{3})=cos\alpha *cos\frac{\pi }{3} -sin\alpha *sin\frac{\pi }{3} =\frac{15}{17} *\frac{1}{2} -(-\frac{5}{17} *\frac{\sqrt{3} }{2})= \\=\frac{15}{34} +\frac{5\sqrt{3} }{34}=\frac{5}{34}*(3+\sqrt{3}).$

2.

$sin(\alpha -\frac{\pi }{4})\ \ \ \ sin\alpha =0,6 \ \ \ \ \ \frac{\pi }{2} < \alpha < \pi.\\ sin^2\alpha +cos^2\alpha =1\\ cos^2\alpha =1-sin^2\alpha =1-0,6^2=1-0,36=0,64\\ cos\alpha =б\sqrt{0,64}=б0,8 \ \ \ \ \ \frac{\pi }{2} < \alpha < \pi \ \ \ \ \Rightarrow\\ cos\alpha =-0,8.\\sin(\alpha -\frac{\pi }{4})=sin\alpha *cos\frac{\pi }{4} -cos\alpha *sin\frac{\pi }{4}=0,6*\frac{\sqrt{2} }{2}-(-0,8*\frac{\sqrt{2} }{2} )=\\ =0,3\sqrt{2}+0,4\sqrt{2}=0,7\sqrt{2}.$