Все 4 задания даю 60 баллов заранее спасибо

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Автор - Trover

$1.;1);1+tg^2alpha=frac1{cos^2alpha}Rightarrowcosalpha=sqrt{frac1{1+tg^2alpha}}\cosalpha=sqrt{frac1{1+frac{225}{64}}}=sqrt{frac1{frac{289}{64}}}=sqrt{frac{64}{289}}=pmfrac8{17}\450^o<alpha<540^o\90^o<alpha<180^oalphaRightarrow cosalpha=-frac8{17}\2);cosalpha=sqrt{1-sin^2alpha}=sqrt{1-0,36}=sqrt{0,64}=pm0,8\cosalpha>0Rightarrowcosalpha=0,8\tgalpha=frac{sinalpha}{cosalpha}=frac{-0,6}{0,8}=-frac34\ctgalpha=frac1{tgalpha}=-frac43$
$3);tgalpha=frac1{ctgalpha}=-frac7{24}\1+tg^2alpha=frac1{cos^2alpha}Rightarrowcosalpha=sqrt{frac1{1+tg^2alpha}}\cosalpha=sqrt{frac1{1+frac{49}{576}}}=sqrt{frac1{frac{625}{576}}}=sqrt{frac{576}{625}}=pmfrac{24}{25}\630^o<alpha<720^oRightarrow270^o<alpha<360^o\cosalpha=frac{24}{25}\sinalpha=sqrt{1-cos^2alpha}=sqrt{1-frac{576}{625}}=sqrt{frac{49}{625}}=pmfrac7{25}\270^o<alpha<360^oRightarrowsinapha=-frac7{25}$

$2.;1);(tg^2alpha-sin^2alpha)ctg^2alpha=tg^2alphacdot ctg^2alpha-sin^2alphacdot ctg^2alpha=\=frac{sin^2alpha}{cos^2alpha}cdotfrac{cos^2alpha}{sin^2alpha}-sin^2alphacdotfrac{cos^2alpha}{sin^2alpha}=1-cos^2alpha=sin^2alpha\2);sin^4alpha+cos^4alpha+2sin^2alphacdotcos^2alpha=(sin^2alpha+cos^2alpha)^2=1^2=1$
$3);frac{1+frac1{tgalpha}+frac1{tg^2alpha}}{1+frac1{ctgalpha}+frac1{ctg^2alpha}}=frac{1+ctgalpha+ctg^2alpha}{frac{ctg^2alpha+ctgalpha+1}{ctg^2alpha}}=frac{ctg^2alpha(1+ctgalpha+ctg^2alpha)}{1+ctgalpha+ctg^2alpha}=ctg^2alpha\\3.;1);frac{sinalpha}{1-cosalpha}+frac{1-cosalpha}{sinalpha}=frac{sin^2alpha+(1-cosalpha)^2}{sinalpha(1-cosalpha)}=$
$=frac{sin^2alpha+1-2cosalpha+cos^2alpha}{sinalpha(1-cosalpha)}=frac{2-2cosalpha}{sinalpha(1-cosalpha)}=frac{2(1-cosalpha)}{sinalpha(1-cosalpha)}=frac2{sinalpha}\2);frac{1-(sinalpha-cosalpha)^2}{1+sin^2alpha-cos^2alpha}=frac{1-sin^2alpha+2sinalphacosalpha-cos^2alpha}{sin^2alpha+sin^2alpha}=$
$=frac{1-(sin^2alpha+cos^2alpha)+2sinalphacosalpha}{2sin^2alpha}=frac{1-1+2sinalphacosalpha}{2sin^2alpha}=frac{2sinalphacosalpha}{2sin^2alpha}=ctgalpha$

$4.;1);tg x+1=sin^2x+cos^2x\tg x+1=1\tg x=0\x=pi k,;kinmathbb{Z}\2);2sin x+cos^2x=2-sin^2x\2sin x=2-sin^2x-cos^2x\2sin x=2-(sin^2x+cos^2x)\2sin x=2-1\2sin x=1\sin x=frac12\x=(-1)^ncdotfracpi6+pi n,;ninmathbb{Z}\3);frac{cos x}{sin x-1}=0$
$begin{cases}cos x=0\sin x-1neq0end{cases}Rightarrowbegin{cases}cos x=0\sin xneq1end{cases}Rightarrowbegin{cases}x=frac{pi}2+pi n\xneqfrac{pi}2+2pi nend{cases}Rightarrow\Rightarrow x=frac{3pi}2+2pi n,;ninmathbb{Z}$

Ответ

Проверено экспертом

Автор - nastasiya98

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