Вычислите пожалуйста, очень надо.

Question 1

Question 2

Решите задачу:

$1)\; \; sina= \sqrt{0,2} \; \; ,\; \; \; cos(2a-\pi )=?\\\\cos(-x)=cosx\; \; \Rightarrow \; \; cos(2a-\pi )=cos(\pi -2a)=-cos2a\\\\-cos2a=-(cos^2a-sin^2a)=-cos^2a+sin^2a$

$sina=\sqrt{0,2}\; \; \to \; \; sin^2a=0,2\\\\sin^2a+cos^2a=1\; \; \to \; \; cos^2a=1-sin^2a=1-0,2=0,8\\\\\\cos(2a-\pi )=-0,8+0,2=-0,6$

$2)\; \; sin(x+45^\circ)+cos(x+45^\circ )=-\sqrt{ \frac{4}{5} }\; \; \to \\\\(sinx\cdot cos45^\circ +cosx\cdot sin45^\circ)+(cosx\cdot cos45^\circ -sinx\cdot cos45^\circ )=- \sqrt{\frac{4}{5} } \\\\2cosx\cdot \frac{\sqrt2}{2}=-\sqrt{\frac{4}{5}}\; \; \to \; \; \sqrt2\cdot cosx=- \sqrt{\frac{4}{5}} \\\\cosx=-\sqrt{ \frac{4}{5} }\cdot \frac{1}{\sqrt2}=- \frac{2}{\sqrt5\cdot \sqrt2} =- \frac{\sqrt2}{\sqrt5}=-\sqrt{\frac{2}{5}}$

$cos3x=4cos^3x-3cosx=4\cdot (-\sqrt{\frac{2}{5}})^3-3\cdot (-\sqrt{\frac{2}{5}})=-4\cdot (\sqrt{\frac{2}{5} })^3+3\cdot \sqrt{\frac{2}{5}}\; ;$

$log_{ \frac{14}{25} }\, |cosx|+log_{ \frac{14}{25} }\, |cos3x|=$

$= log_{\frac{14}{25} }\Big |-\sqrt{\frac{2}{5}}\, \Big |+log_ {\frac{14}{25} }\, \Big |3\sqrt{\frac{2}{5}}-4(\sqrt{\frac{2}{5}})^3\Big |=\\\\=log_{\frac{14}{25}}\; \sqrt{\frac{2}{5}}+log_{\frac{14}{25}}\; \Big (\sqrt{\frac{2}{5}}\cdot (3-\frac{4\cdot 2}{5})\Big )=$

$=log_{\frac{14}{25}}\, \sqrt{\frac{2}{5}}+log_{\frac{14}{25}}\, \Big (\sqrt{\frac{2}{5}}\cdot \frac{7}{5}\Big ) =2\cdot log_{\frac{14}{25}}\sqrt{\frac{2}{5}}+log_{\frac{14}{25} } \frac{7}{5}=$

$=2\cdot \frac{log_5\sqrt{2/5}}{log_5(14/25)} + log_{\frac{14}{25}}\, \frac{7}{5} =2\cdot \frac{log_5\, \sqrt2-log_5\, \sqrt5}{log_5\, (7\cdot 2)-log_5\, 25} +log_{ \frac{14}{25} } \frac{7}{5} =$

$= 2\cdot \frac{\frac{1}{2}log_52-\frac{1}{2}}{log_57+log_52-log_55^2} + \frac{log_57-log_55}{log_57+log_52-log_55^2} = \frac{log_52-1+log_57-1}{log_57+log_52-2} =\\\\= \frac{log_52+log_57-2}{log_57+log_52-2} =1$

Question 3

Смотри во вложении.