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Question 1

Question 2

Решите задачу:

$1)\; \; \alpha =-2\pi +\frac{\pi}{4}=-\frac{7\pi}{4}\\\\sin(-\frac{7\pi}{4})=sin(-2\pi +\frac{\pi}{4})=sin \frac{\pi }{4}= \frac{\sqrt2}{2}\\\\cos(-\frac{7\pi }{4})=cos\frac{\pi}{4}= \frac{\sqrt2}{2}\\\\tg(- \frac{7\pi }{4})=tg \frac{\pi}{4}=1 \\\\ctg( \frac{7\pi}{4})=ctg \frac{\pi }{4}=1$

$2)\; \; \frac{cos(-245^\circ )\cdot ctg135^\circ }{tg189^\circ } = \frac{cos245^\circ \cdot ctg135^\circ}{tg189^\circ} = \frac{(-)\cdot (-)}{(+)}= \frac{(+)}{(+)}=(+)$

$3)\; \; sina=0,8\; ,\\\\ \frac{\pi }{2}\ \textless \ a\ \textless \ \pi \; \; \to \; \; cosa\ \textless \ 0\; ,\; \; ctga\ \textless \ 0\\\\cosa=-\sqrt{1-sin^2a}=-\sqrt{1-0,64}=-0,6\\\\1+ctg^2a= \frac{1}{sin^2a} \; \; \to \; \; ctg^2a= \frac{1}{sin^2a}-1=\frac{1}{0,64}-1= \frac{0,36}{0,64}=(\frac{0,6}{0,8})^2\\\\ctga=-\sqrt{(\frac{0,6}{0,8})^2}=-\frac{0,6}{0,8}=-\frac{3}{4}\\\\1-ctga+\frac{1}{cosa}=1-(-\frac{3}{4})+\frac{1}{-0,6}=1+\frac{3}{4}-\frac{10}{6}= \frac{1}{12}$

$4)\; \; \frac{cosa+ctga}{ctga} = \frac{cosa}{ctga}+ \frac{ctga}{ctga}= \frac{cosa}{\frac{cosa}{sina}}+1=sina+1$

$5)\; \; sin20\cdot sin40\cdot sin80=sin20\cdot \frac{1}{2}(cos(80-40)-cos(80+40))=\\\\= \frac{1}{2}sin20\cdot (cos40-cos120)= \frac{1}{2}sin20\cdot (cos40-cos(90+30))=\\\\= \frac{1}{2}\cdot \frac{1}{2}\cdot (sin(20-40)+sin(20+40))+\frac{1}{2}\cdot sin20\cdot sin30= \\\\= \frac{1}{4}\cdot sin(-20)+ \frac{1}{4}\cdot sin60+\frac{1}{2}\cdot sin20\cdot \frac{1}{2}=\\\\= - \frac{1}{4}\cdot sin20+\frac{1}{4} \cdot \frac{\sqrt3}{2}+ \frac{1}{4}\cdot sin20= \frac{\sqrt3}{8}$