1) Найдите значение ctg(3п/2 + альфа), если ctg альфа = 10/11 2) Найдите значение...

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мохинсан_zn · Answer 1 · 2018-03-16T19:30:43+0000

411
$ctg\alpha= \frac{10}{11};\\ ctg( \frac{3\pi}{2}+\alpha )= \frac{\cos( \frac{3\pi}{2}+\alpha) }{\sin( \frac{3\pi}{2}+\alpha )}= \frac{\cos \frac{3\pi}{2}\cos\alpha-\sin\frac{3\pi}{2}\sin\alpha }{\sin\frac{3\pi}{2}\cos\alpha+\sin\alpha\cos\frac{3\pi}{2}}=\\ |\cos\frac{3\pi}{2}=0;\ \ \ \sin\frac{3\pi}{2}=-1|\\ = \frac{0\cdot\cos\alpha-(-1)\cdot\sin\alpha}{(-1)\cdot\cos\alpha+\sin\alpha\cdot0}= \frac{\sin\alpha}{-\cos\alpha}=-tg\alha=- \frac{1}{ctg\alpha}=- \frac{1}{ \frac{10}{11} }= -\frac{11}{10}=\\$ $=-1 \frac{1}{10}=-1,1$
413
tga(\alpha+180^0\cdotk)=tg\alpha, k\in Z\\ ctg(\alpha+\pi k)=ctg\alpha,==>ctg(\alpha+180^0\cdotk)=ctg\alpha k\in Z\\ =2tg(15^0+6\cdot180^0)+ctg(15^0+5\cdot180^0)-tg(-15^0-180^0)=\\ =2tg15^0+ctg15^0-tg(-15^0)=\\ |tg(-alpha)= \frac{\sin(-\alpha)}{\cos(-\alpha)}= \frac{-\sin\alpha}{\cos\alpha}=-tg\alpha|\\ 2t \frac{x}{y} 15^0+ \frac{1}{tg15^0}+tg15^0=\frac{3tg^215^0+1}{tg15^0}=\frac{3(2-\sqrt{3})^2+1}{2-\sqrt{3}}= \\ " alt="tg15^0=2-\sqrt{3};\\ 2tg1095^0+ctg975^0-tg(-195)=\\ |tg(\alpha+\pi k)=tg\alpha,==>tga(\alpha+180^0\cdotk)=tg\alpha, k\in Z\\ ctg(\alpha+\pi k)=ctg\alpha,==>ctg(\alpha+180^0\cdotk)=ctg\alpha k\in Z\\ =2tg(15^0+6\cdot180^0)+ctg(15^0+5\cdot180^0)-tg(-15^0-180^0)=\\ =2tg15^0+ctg15^0-tg(-15^0)=\\ |tg(-alpha)= \frac{\sin(-\alpha)}{\cos(-\alpha)}= \frac{-\sin\alpha}{\cos\alpha}=-tg\alpha|\\ 2t \frac{x}{y} 15^0+ \frac{1}{tg15^0}+tg15^0=\frac{3tg^215^0+1}{tg15^0}=\frac{3(2-\sqrt{3})^2+1}{2-\sqrt{3}}= \\ " align="absmiddle" class="latex-formula"> $= \frac{3(2-\sqrt{3})^2+1}{2-\sqrt{3}}= \frac{12-12\sqrt{3}+9+1}{2-\sqrt{3}}= \\ \frac{(22-12\sqrt{3})\cdot(2+\sqrt{3})}{(2-\sqrt{3})\cdot(2+\sqrt{3})}= \frac{44-24\sqrt{3}+22\sqrt{3}-12\cdot3}{4-3}= \frac{44-36-2\sqrt{3}}{1}=8-2\sqrt{3};$