Question

решите срочно пожалуйста дам 100 баллов
под В2

Answer 1

$1)\; \; \frac{sin\frac{\pi}{10}\, cos\frac{3\pi }{20}+cos\frac{\pi}{10}\, sin\frac{3\pi}{20}}{cos15^\circ }=\frac{sin(\frac{\pi}{10}+\frac{3\pi}{20})}{cos(45^\circ -30^\circ )}=\frac{sin\frac{5\pi }{20}}{cos45^\circ \, cos30^\circ +sin45^\circ \, sin30^\circ }=\\\\= \frac{sin\frac{\pi}{4}}{\frac{\sqrt2}{2}\cdot \frac{\sqrt3}{2}+\frac{\sqrt2}{2}\cdot \frac{1}{2}}=\frac{\frac{\sqrt2}{2}}{\frac{\sqrt2(\sqrt3+1)}{4}}=\frac{2}{\sqrt3+1}=\frac{2\, (\sqrt3-1)}{(\sqrt3+1)(\sqrt3-1)}=\frac{2\, (\sqrt3-1)}{2}=\sqrt3-1$

$2)\; \; ctga=-\frac{1}{\sqrt3}\\\\90^\circ \ \textless \ a\ \textless \ 180^\circ \; \; \; \Rightarrow \; \; cosa\ \textless \ 0\; ,\; \; sina\ \textgreater \ 0\; .\\\\1+ctg^2a=\frac{1}{sin^2a}\; \; \to \; \; sin^2a=\frac{1}{1+ctg^2a}=\frac{1}{1+\frac{1}{3}}=\frac{3}{4}\; ,\; sina=+\frac{\sqrt3}{2}\ \textgreater \ 0\\\\\cos^2a=1-sin^2a=1-\frac{3}{4}=\frac{1}{4}\; ,\; \; cosa=-\frac{1}{2}\ \textless \ 0\\\\\\cos(\frac{\pi }{3}-a)=cos\frac{\pi}{3}\cdot cosa+sin\frac{\pi}{3}\cdot sina=\frac{1}{2}\cdot (-\frac{1}{2})+\frac{\sqrt3}{2}\cdot \frac{\sqrt3}{2}=\\\\=-\frac{1}{4}+\frac{3}{4}=\frac{2}{4}=\frac{1}{2}$