Question

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Answer 1

$1)\; \; cos(-45^\circ )=cos45^\circ =\frac{\sqrt2}{2}\\\\sin405^\circ =sin(360^\circ+45^\circ )=sin45^\circ =\frac{\sqrt2}{2}\\\\cos900^\circ =cos(720^\circ +180^\circ )=cos180^\circ =-1\\\\tg(-\frac{\pi}{3})=-tg\frac{\pi}{3}=-\sqrt3\\\\ctg(-\frac{7\pi }{3})=-ctg(2\pi +\frac{\pi}{3})=-ctg\frac{\pi}{3}=-\frac{\sqrt3}{3}\\\\\frac{cos(-45^\circ )\cdot sin405^\circ -cos900^\circ }{tg(-\frac{\pi}{3})\cdot ctg(-\frac{7\pi}{3})}=\frac{\frac{\sqrt2}{2}\cdot \frac{\sqrt2}{2}+1}{\sqrt3\cdot \frac{\sqrt3}{3}}=\frac{\frac{1}{2}+1}{1}=1,5$

$2)\; \; 1-sin\frac{\pi}{6}+sin^2\frac{\pi}{6}-sin^3\frac{\pi}{6}+,,,=1-\frac{1}{2}+(\frac{1}{2})^2-(\frac{1}{2})^3+...=\\\\=\Big (\underbrace {1+(\frac{1}{2})^2+(\frac{1}{2})^4+...}_{b_1=1\; ,\; q=\frac{1}{4}}\Big )-\Big (\, \underbrace {\frac{1}{2}+(\frac{1}{2})^3+(\frac{1}{2})^5+...}_{b_1=1/2\; ,\; q=\frac{1}{4}}\Big )=\\\\=\frac{1}{1-(\frac{1}{2})^2}-\frac{\frac{1}{2}}{1-(\frac{1}{2})^2}=\frac{1-\frac{1}{2}}{1-\frac{1}{4}}=\frac{\frac{1}{2}}{\frac{3}{4}}=\frac{2}{3}$