Question

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Answer 1

$\displaystyle\bf\\Cos\alpha =-\frac{5}{13} \ \ \ ; \ \ \ Sin\beta =\frac{3}{5} \\\\\\Sin\alpha =\sqrt{1-Cos^{2}\alpha } =\sqrt{1-\Big(-\frac{5}{13}\Big)^{2} }=\sqrt{1-\frac{25}{169} }=\\\\\\=\sqrt{\frac{144}{169} }=\sqrt{\Big(\frac{12}{13}\Big)^{2} } =\frac{12}{13} \\\\\\Cos\beta =-\sqrt{1-Sin^{2}\beta } =-\sqrt{1-\Big(\frac{3}{5}\Big)^{2} }=-\sqrt{1-\frac{9}{25} } =\\\\\\=-\sqrt{\frac{16}{25} } =-\sqrt{\Big(\frac{4}{5} \Big)^{2} }=-\frac{4}{5}$

$\displaystyle\bf\\Cos(\alpha +\beta) =Cos\alpha Cos\beta -Sin\alpha Sin\beta =-\frac{5}{13}\cdot \Big(-\frac{4}{5} \Big)-\frac{12}{13}\cdot\frac{3}{5}=\\\\\\=\frac{20}{65} -\frac{36}{65} =-\frac{16}{65}$