от mail_dinko » 01 Мар 2012, 21:50
[tex]\alpha\in (\frac{\pi}{2}; \pi)[/tex]--->втори квадрант--> само синусът е положителен
[tex]cotg(\alpha - \frac{ \pi}{4})=?[/tex]
[tex]\fbox {sin\alpha =\frac13}[/tex]
[tex]sin^2 \alpha +cos^2 \alpha =1[/tex]
[tex](\frac13)^2+cos^2 \alpha =1[/tex]
[tex]\frac19+cos^2 \alpha =1|.9[/tex]
[tex]1+9cos^2 \alpha =9[/tex]
[tex]9cos^2 \alpha = 9-1[/tex]
[tex]9cos^2 \alpha = 8|:9[/tex]
[tex]cos^2 \alpha = \frac89[/tex]
[tex]cos \alpha = \pm \sqrt {\frac89}[/tex]
[tex]cos \alpha = \pm \frac {\sqrt {8}}{\sqrt {9}}[/tex]
[tex]cos \alpha = \pm \frac {2\sqrt {2}}{3}[/tex]
Вземаме отрицателната стойност, заради това, че ъгълът е във втори квадрант.
[tex]\fbox {cos \alpha = - \frac {2\sqrt {2}}{3}}[/tex]
[tex]tg \alpha = \frac {sin \alpha} {cos \alpha}[/tex]
[tex]tg \alpha = \frac {\frac13}{- \frac {2\sqrt {2}}{3}}[/tex]
[tex]tg \alpha = \frac{1}{3} : (- \frac {2\sqrt {2}}{3} )[/tex]
[tex]tg \alpha =-\frac{1}{\cancel{3}} . \frac {\cancel{3}}{2 \sqrt {2}}[/tex]
[tex]\fbox{tg \alpha = - \frac {1}{2 \sqrt{2}}}= - \frac {1}{2 \sqrt{2}} . \frac {\sqrt {2}}{\sqrt {2}}=\fbox{tg \alpha = - \frac {\sqrt{2}}{4}}[/tex]
[tex]tg \alpha . cotg \alpha =1[/tex]
[tex]cotg \alpha = \frac{1}{tg \alpha}=\frac {1}{- \frac {1}{2 \sqrt{2}}}=-2 \sqrt{2}[/tex]
[tex]\fbox {cotg \alpha = -2 \sqrt{2}}[/tex]
[tex]\frac { \pi}{4}= \frac {180 ^\circ }{4}=45^\circ[/tex]
[tex]cotg(\alpha - 45^\circ )=\frac{cotg \alpha cotg 45^\circ +1}{cotg 45^\circ -cotg \alpha} =\frac{-2\sqrt{2}+1 }{1-(-2 \sqrt {2}) }=\frac{1-2 \sqrt{2}}{1+2\sqrt{2} }.\frac {1-2\sqrt {2}}{1-2\sqrt {2}}=\frac{1-4\sqrt{2}+8}{ 1-8}=\frac{9-4\sqrt{2}}{-7 }[/tex]
[tex]\fbox {cotg (\alpha-45^\circ)=\frac{4\sqrt {2}-9}{7 }[/tex]