1)Вычислите косинусы углов: 2) Вычислить: sin(a-β),если cosa=-

$cos75^0=cos(30^0+45^0)=cos30cos45-sin30sin45=\\\\=\frac{\sqrt3}{2}\cdot \frac{\sqrt2}{2}-\frac{1}{2}\cdot \frac{\sqrt2}{2}=\frac{\sqrt2(\sqrt3-1)}{4}\\\\cos15^0=cos(45^0-30^0)=cos45cos30+sin45sin30=\\\\=\frac{\sqrt2}{2}\cdot \frac{\sqrt3}{2}+\frac{\sqrt2}{2}\cdot \frac{1}{2}=\frac{\sqrt2(\sqrt3+1)}{4}$

0 \\\\sin \alpha =-\sqrt{1-cos^2 \alpha }=-\sqrt{1-\frac{16}{25}}=-\sqrt{\frac{9}{25}}=-\frac{3}{5}\\\\cos \beta =+\sqrt{1-sin^2 \beta }=\sqrt{1-\frac{24^2}{25^2}}=\sqrt{\frac{49}{625}}=\frac{7}{25}\\\\sin( \alpha - \beta )=sin \alpha \cdot cos \beta -sin \beta \cdot cos \alpha =" alt="cos \alpha =-\frac{4}{5};\; \; \pi < \alpha <\frac{3\pi }{2}\; \; \to \; \; sin \alpha <0\\\\sin \beta =-\frac{24}{25};\; \; \frac{3\pi }{2}< \beta <2\pi\; \; \to \; \; cos \beta >0 \\\\sin \alpha =-\sqrt{1-cos^2 \alpha }=-\sqrt{1-\frac{16}{25}}=-\sqrt{\frac{9}{25}}=-\frac{3}{5}\\\\cos \beta =+\sqrt{1-sin^2 \beta }=\sqrt{1-\frac{24^2}{25^2}}=\sqrt{\frac{49}{625}}=\frac{7}{25}\\\\sin( \alpha - \beta )=sin \alpha \cdot cos \beta -sin \beta \cdot cos \alpha =" align="absmiddle" class="latex-formula">

$=-\frac{3}{5}\cdot \frac{7}{25}-(-\frac{4}{5})\cdot (-\frac{24}{25})=\frac{-21-96}{125}=-\frac{117}{125}$