Все 4 задания даю 60 баллов заранее спасибо

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Дано ответов: 2

Trover_zn · Answer 1 · 2018-04-02T21:00:49+0000

0\Rightarrow\cos\alpha=0,8\\tg\alpha=\frac{\sin\alpha}{\cos\alpha}=\frac{-0,6}{0,8}=-\frac34\\ctg\alpha=\frac1{tg\alpha}=-\frac43" alt="1.\;1)\;1+tg^2\alpha=\frac1{\cos^2\alpha}\Rightarrow\cos\alpha=\sqrt{\frac1{1+tg^2\alpha}}\\\cos\alpha=\sqrt{\frac1{1+\frac{225}{64}}}=\sqrt{\frac1{\frac{289}{64}}}=\sqrt{\frac{64}{289}}=\pm\frac8{17}\\450^o<\alpha<540^o\\90^o<\alpha<180^o\alpha\Rightarrow \cos\alpha=-\frac8{17}\\2)\;\cos\alpha=\sqrt{1-\sin^2\alpha}=\sqrt{1-0,36}=\sqrt{0,64}=\pm0,8\\\cos\alpha>0\Rightarrow\cos\alpha=0,8\\tg\alpha=\frac{\sin\alpha}{\cos\alpha}=\frac{-0,6}{0,8}=-\frac34\\ctg\alpha=\frac1{tg\alpha}=-\frac43" align="absmiddle" class="latex-formula">
$3)\;tg\alpha=\frac1{ctg\alpha}=-\frac7{24}\\1+tg^2\alpha=\frac1{\cos^2\alpha}\Rightarrow\cos\alpha=\sqrt{\frac1{1+tg^2\alpha}}\\\cos\alpha=\sqrt{\frac1{1+\frac{49}{576}}}=\sqrt{\frac1{\frac{625}{576}}}=\sqrt{\frac{576}{625}}=\pm\frac{24}{25}\\\630^o<\alpha<720^o\Rightarrow270^o<\alpha<360^o\\\cos\alpha=\frac{24}{25}\\\sin\alpha=\sqrt{1-\cos^2\alpha}=\sqrt{1-\frac{576}{625}}=\sqrt{\frac{49}{625}}=\pm\frac7{25}\\270^o<\alpha<360^o\Rightarrow\sin\apha=-\frac7{25}$

$2.\;1)\;(tg^2\alpha-\sin^2\alpha)ctg^2\alpha=tg^2\alpha\cdot ctg^2\alpha-\sin^2\alpha\cdot ctg^2\alpha=\\=\frac{\sin^2\alpha}{\cos^2\alpha}\cdot\frac{\cos^2\alpha}{\sin^2\alpha}-\sin^2\alpha\cdot\frac{\cos^2\alpha}{\sin^2\alpha}=1-\cos^2\alpha=\sin^2\alpha\\2)\;\sin^4\alpha+\cos^4\alpha+2\sin^2\alpha\cdot\cos^2\alpha=(\sin^2\alpha+\cos^2\alpha)^2=1^2=1$
$3)\;\frac{1+\frac1{tg\alpha}+\frac1{tg^2\alpha}}{1+\frac1{ctg\alpha}+\frac1{ctg^2\alpha}}=\frac{1+ctg\alpha+ctg^2\alpha}{\frac{ctg^2\alpha+ctg\alpha+1}{ctg^2\alpha}}=\frac{ctg^2\alpha(1+ctg\alpha+ctg^2\alpha)}{1+ctg\alpha+ctg^2\alpha}=ctg^2\alpha\\\\3.\;1)\;\frac{\sin\alpha}{1-\cos\alpha}+\frac{1-\cos\alpha}{\sin\alpha}=\frac{\sin^2\alpha+(1-\cos\alpha)^2}{\sin\alpha(1-\cos\alpha)}=$
$=\frac{\sin^2\alpha+1-2\cos\alpha+\cos^2\alpha}{\sin\alpha(1-\cos\alpha)}=\frac{2-2\cos\alpha}{\sin\alpha(1-\cos\alpha)}=\frac{2(1-\cos\alpha)}{\sin\alpha(1-\cos\alpha)}=\frac2{\sin\alpha}\\2)\;\frac{1-(\sin\alpha-\cos\alpha)^2}{1+\sin^2\alpha-\cos^2\alpha}=\frac{1-\sin^2\alpha+2\sin\alpha\cos\alpha-\cos^2\alpha}{\sin^2\alpha+\sin^2\alpha}=$
$=\frac{1-(\sin^2\alpha+\cos^2\alpha)+2\sin\alpha\cos\alpha}{2\sin^2\alpha}=\frac{1-1+2\sin\alpha\cos\alpha}{2\sin^2\alpha}=\frac{2\sin\alpha\cos\alpha}{2\sin^2\alpha}=ctg\alpha$

$4.\;1)\;tg x+1=\sin^2x+\cos^2x\\tg x+1=1\\tg x=0\\x=\pi k,\;k\in\mathbb{Z}\\2)\;2\sin x+\cos^2x=2-\sin^2x\\2\sin x=2-\sin^2x-\cos^2x\\2\sin x=2-(\sin^2x+\cos^2x)\\2\sin x=2-1\\2\sin x=1\\\sin x=\frac12\\x=(-1)^n\cdot\frac\pi6+\pi n,\;n\in\mathbb{Z}\\3)\;\frac{\cos x}{\sin x-1}=0$
$\begin{cases}\cos x=0\\\sin x-1\neq0\end{cases}\Rightarrow\begin{cases}\cos x=0\\\sin x\neq1\end{cases}\Rightarrow\begin{cases}x=\frac{\pi}2+\pi n\\x\neq\frac{\pi}2+2\pi n\end{cases}\Rightarrow\\\Rightarrow x=\frac{3\pi}2+2\pi n,\;n\in\mathbb{Z}$