Предмет: Математика
Автор: Orestes1300
Вопрос задан 2 года назад

Все 4 задания , даю 60баллов большое спасибо

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Ответы

Ответ дал: Аноним

$1.\;1)\;\cos\alpha=-0,6\\\sin\alpha=\sqrt{1-\cos^2\alpha}=\sqrt{1-0,36}=\sqrt{0,64}=\pm0,8\\90^o<\alpha<180^o\Rightarrow\sin\alpha=+0,8\\tg\alpha=\frac{\sin\alpha}{\cos\alpha}=\frac{0,8}{-0,6}=-\frac43=-1\frac13\\2)\;1+tg^2\alpha=\frac1{\cos^2\alpha}\Rightarrow\cos\alpha=\sqrt{\frac1{1+tg^2\alpha}}=\sqrt{\frac1{1+\frac9{16}}}=\sqrt{\frac1{\frac{25}{16}}}=\sqrt{\frac{16}{25}}=\\=\pm\frac45\\270^o<\alpha<360^o\Rightarrow\cos\alpha=+\frac45$
$tg\alpha=\frac{\sin\alpha}{\cos\alpha}\Rightarrow\sin\alpha=tg\alpha\cdot\cos\alpha=-\frac34\cdot\left(-\frac45\right)=-\frac35\\3)\;\cos\alpha=\sqrt{1-\sin^2\alpha}=\sqrt{1-\frac{144}{169}}=\sqrt{\frac{25}{169}}=\pm\frac5{13}\\\frac\pi2<\alpha<\pi\Rightarrow\cos\alpha=-\frac5{13}\\tg\alpha=\frac{\sin\alpha}{\cos\alpha}=\frac{12}{13}:\left(-\frac5{13}\right)=-\frac{12}{13}\cdot\frac{13}5=-\frac{12}5\\ctg\alpha=\frac1{tg\alpha}=-\frac5{12}$

$2.\;1)\;\left(\frac{\sin\alpha}{tg\alpha}\right)^2+\left(\frac{\cos\alpha}{ctg\alpha}\right)^2-\sin^2\alpha=\\=\left({\sin\alpha}:\frac{\sin\alpha}{\cos\alpha}\right)^2+\left({\cos\alpha}:\frac{\cos\alpha}{\sin\alpha}\right)^2+\sin^2\alpha=\\=\left({\sin\alpha}\cdot\frac{\cos\alpha}{\sin\alpha}\right)^2+\left({\cos\alpha}\cdot\frac{\sin\alpha}{\cos\alpha}\right)^2-\sin^2\alpha=\\=(\cos\alpha)^2+(\sin\alpha)^2-\sin^2\alpha=\cos^2\alpha+\sin^2\alpha-\sin^\alpha=\cos^2\alpha$
$2)\;(1+ctg^2\alpha)(1-\sin^2\alpha)=1-\sin^2\alpha+ctg^2\alpha-\sin^2\alpha ctg^2\alpha=\\=1-\sin^2\alpha-\sin^2\alpha\cdot\frac{\cos^2\alpha}{\sin^2\alpha}+ctg^2\alpha=1-\sin^2\alpha-\cos^2\alpha+ctg^2\alpha=\\=1-(\sin^2\alpha+\cos^2\alpha)+cth\alpha=1-1+ctg^2\alpha=ctg^2\alpha$
$3)\;\frac{1+tg\alpha+tg^2\alpha}{1+ctg\alpha+ctg^2\alpha}=\frac{1+tg\alpha+tg^2\alpha}{1+\frac1{tg\alpha}+\frac1{tg^2\alpha}}=\frac{1+tg\alpha+tg^2\alpha}{\frac{tg^2\alpha+tg\alpha+1}{tg^2\alpha}}=\frac{tg^2\alpha(1+tg\alpha+tg^2)}{1+tg\alpha+tg^2}=tg^2\alpha$

$3.\;1)\;\frac{2\sin^2\alpha-1}{\sin\alpha-\cos\alpha}=\frac{1-2\sin^2\alpha}{\cos\alpha-\sin\alpha}=\frac{\cos2\alpha}{\cos\alpha-\sin\alpha}=\frac{\cos^2\alpha-\sin^2\alpha}{\cos\alpha-\sin\alpha}=\\=\frac{(\cos\alpha-\sin\alpha)(\cos\alpha+\sin\alpha)}{\cos\alpha-\sin\alpha}=\cos\alpha+\sin\alpha$
$2)\;(a\sin\alpha+b\cos\alpha)^2+(a\cos\alpha-b\sin\alpha)^2=\\=a^2\sin^2\alpha+2ab\sin\alpha\cos\alpha+b^2\cos^2\alpha+a^2\cos^2\alpha-2ab\sin\alpha\cos\alpha+\\+b^2\sin^2\alpha=\sin^2\alpha(a^2+b^2)+\cos^2\alpha(a^2+b^2)=\\=(a^2+b^2)(\sin^2\alpha+\cos^2\alpha)=a^2+b^2$

$4.\;1)\;\sin^2x-\cos^2x+1=0\\\sin^2x-\cos^2x+\sin^2x+\cos^2x=0\\2\sin^2x=0\\\sin^2x=0\\\sin x=0\\x=\pi n,\;n\in\mathbb{Z}\\2)\;tgx\cdot ctgx=\cos x\\\frac{\sin x}{\cos x}\cdot\frac{\cos x}{\sin x}=\cos x\\\cos x=1\\x=2\pi n,\;n\in\mathbb{Z}$
$3)\;\sin x-\sin^2x+tg45^o=\cos^2x+\cos45^o\\\sin x=\cos^2x+\sin^2x+\cos45^o-tg45^o\\\sin x=1+\frac1{\sqrt2}-1\\\sin x=\frac1{\sqrt2}\\x=(-1)^n\cdot\frac\pi4+\pi n,\;n\in\mathbb{Z}$

Ответ дал: nastasiya98