Question

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Тригонометрия

Answer 1

Объяснение:

178.

$1)\ cos\alpha =\frac{4}{5}\ \ \ \ \frac{3\pi }{2}< \alpha < 2\pi \ \ \ \ \ sin\beta =-\frac{3}{5}\ \ \ \ \ \pi < \beta < \frac{3\pi }{2} .\\sin(\alpha +\beta )=sin\alpha *cos\beta +cos\alpha *sin\beta \\sin^2\alpha +cos^2\alpha =1\\sin^2\alpha =1-cos^2\alpha =1-(\frac{4}{5})^2=1-\frac{16}{25} =\frac{9}{25} .\\ sin\alpha =б\sqrt{\frac{9}{25} }=б\frac{3}{5}\ \ \ \ \ \frac{3\pi }{2}< \alpha < 2\pi \ \ \ \ \ \Rightarrow\\ sin\alpha =-\frac{3}{5}.$

$sin^2\beta +cos^2\beta =1\\ cos^2\beta =1-sin^2\beta =1-(-\frac{3}{5})^2=1-\frac{9}{25}=\frac{16}{25}.\\ cos\beta = б\sqrt{\frac{16}{25} } = б\frac{4}{5}\ \ \ \ \ \pi < \beta< \frac{3\pi }{2} \ \ \ \ \Rightarrow\\ cos\beta =-\frac{4}{5}.\\ sin(\alpha +\beta )=-\frac{3}{5}*(-\frac{4}{5})+ \frac{4}{5}*(-\frac{3}{5}) =\frac{12}{25}-\frac{12}{25}=0 .\\ sin(\alpha -\beta )=-\frac{3}{5}*(-\frac{4}{5})- \frac{4}{5}*(-\frac{3}{5}) =\frac{12}{25}+\frac{12}{25}=\frac{24}{25}.$

$2)\ sin\alpha =\frac{8}{17} \ \ \ \ \frac{\pi }{2}< \alpha < \pi \ \ \ \ \ cos\beta =\frac{3}{5}\ \ \ \ \ \ \frac{3\pi }{2}< \beta < 2\pi . \\ cos(\alpha +\beta )=cos\alpha *cos\beta -sin\alpha *sin\beta \\cos^2\alpha =1-(\frac{8}{17})^2=1-\frac{64}{289}=\frac{225}{289}\\ cos\alpha =б\sqrt{\frac{225}{289} } =б\frac{15}{17} \ \ \ \ \frac{\pi }{2}, \alpha < \pi\ \ \ \ \Rightarrow\\ cos\alpha =-\frac{15}{17}.$

$sin^2\beta =1-(\frac{3}{5} )^2=1-\frac{9}{25} =\frac{16}{25} \\sin\beta =б\sqrt{\frac{16}{25} } =б\frac{4}{5} \ \ \ \ \frac{3\pi }{2} < \beta < 2\pi \ \ \ \ \Rightarrow\\sin\beta =-\frac{4}{5}.$

$cos(\alpha +\beta )=(-\frac{15}{17} )*\frac{3}{5} -\frac{8}{17}*(-\frac{4}{5} )=-\frac{45}{85} +\frac{32}{85}=-\frac{13}{85}.\\ cos(\alpha -\beta )=(-\frac{15}{17} )*\frac{3}{5} +\frac{8}{17}*(-\frac{4}{5} )=-\frac{45}{85} -\frac{32}{85}=-\frac{77}{85} .$

$3)\ tg\alpha =-\frac{3}{4} \ \ \ \ \ \frac{\pi }{2} <\alpha <\pi \\sin(\frac{\pi }{4} +\alpha )=sin\frac{\pi }{4}*cos\alpha +cos\frac{\pi }{4}*sin\alpha=\frac{\sqrt{2} }{2} *cos\alpha +\frac{\sqrt{2} }{2}*sin\alpha =\frac{\sqrt{2} }{2}*(sin\alpha +cos\alpha ) \\ (tg\alpha )^2=(-\frac{3}{4})^2\\ tg^2\alpha =\frac{9}{16} \\ \frac{sin^2\alpha }{cos^2\alpha } =\frac{9}{16} \\16*sin^2\alpha =9*cos^2\alpha$

$16*sin^2\alpha +16*cos^2\alpha =9*cos^2\alpha +16*cos^2\alpha \\16*(sin^2\alpha +cos^2\alpha )=25*cos^2\alpha \\25*cos^2\alpha =16\ |:25\\cos^2\alpha =\frac{16}{25} \\cos\alpha =б\sqrt{\frac{16}{25} }= б\frac{4}{5} \ \ \ \ \ \frac{\pi }{2}<\alpha <\pi \ \ \ \ \Rightarrow\\ cos\alpha =-\frac{4}{5} .\\\frac{sin\alpha }{-\frac{4}{5} } =-\frac{3}{4}\\ sin\alpha =(-\frac{4}{5})*(-\frac{3}{4})=\frac{3}{5}.$

$sin(\frac{\pi }{4}+\alpha )=\frac{\sqrt{2} }{2}*(\frac{3}{5}+(-\frac{4}{5}))=\frac{\sqrt{2} }{2}*(\frac{3}{5}-\frac{4}{5}) =-\frac{\sqrt{2} }{10}.\\ sin(\frac{\pi }{4}-\alpha )= \frac{\sqrt{2} }{2}*(-\frac{4}{5}-\frac{3}{5})=\frac{\sqrt{2} }{2}*\frac{-7}{5}=-\frac{7\sqrt{2} }{10}.$