459 реши
Сверь числа и знаки с фото.
459 По значению одной из тригонометрических функций (sin α, cos α, tg α, ctg α) найти значения остальных трёх:
- \(\cos\alpha=\frac{5}{13}\) и \(\frac{3\pi}{2}<\alpha<2\pi\);
- \(\sin\alpha=0{,}8\) и \(\frac{\pi}{2}<\alpha<\pi\);
- \(\mathrm{tg}\alpha=\frac{15}{8}\) и \(\pi<\alpha<\frac{3\pi}{2}\);
- \(\mathrm{ctg}\alpha=-3\) и \(\frac{3\pi}{2}<\alpha<2\pi\);
- \(\cos\alpha=0{,}8\) и \(0<\alpha<\frac{\pi}{2}\);
- \(\sin\alpha=-\frac{5}{13}\) и \(\frac{3\pi}{2}<\alpha<2\pi\);
- \(\mathrm{tg}\alpha=-2{,}4\) и \(\frac{\pi}{2}<\alpha<\pi\);
- \(\mathrm{ctg}\alpha=\frac{7}{24}\) и \(\pi<\alpha<\frac{3\pi}{2}\).
Краткий ответ — все значения в таблице. В столбцах стоят \(\sin\alpha,\ \cos\alpha,\ \mathrm{tg}\,\alpha,\ \mathrm{ctg}\,\alpha\).
| № | Дано | \(\sin\alpha\) | \(\cos\alpha\) | \(\mathrm{tg}\,\alpha\) | \(\mathrm{ctg}\,\alpha\) |
|---|---|---|---|---|---|
| 1 | \(\cos\alpha=\frac{5}{13},\ \frac{3\pi}{2}<\alpha<2\pi\) | \(-\frac{12}{13}\) | \(\frac{5}{13}\) | \(-\frac{12}{5}\) | \(-\frac{5}{12}\) |
| 2 | \(\sin\alpha=0{,}8,\ \frac{\pi}{2}<\alpha<\pi\) | \(\frac45\) | \(-\frac35\) | \(-\frac43\) | \(-\frac34\) |
| 3 | \(\mathrm{tg}\alpha=\frac{15}{8},\ \pi<\alpha<\frac{3\pi}{2}\) | \(-\frac{15}{17}\) | \(-\frac{8}{17}\) | \(\frac{15}{8}\) | \(\frac{8}{15}\) |
| 4 | \(\mathrm{ctg}\alpha=-3,\ \frac{3\pi}{2}<\alpha<2\pi\) | \(-\frac{\sqrt{10}}{10}\) | \(\frac{3\sqrt{10}}{10}\) | \(-\frac13\) | \(-3\) |
| 5 | \(\cos\alpha=0{,}8,\ 0<\alpha<\frac{\pi}{2}\) | \(\frac35\) | \(\frac45\) | \(\frac34\) | \(\frac43\) |
| 6 | \(\sin\alpha=-\frac{5}{13},\ \frac{3\pi}{2}<\alpha<2\pi\) | \(-\frac{5}{13}\) | \(\frac{12}{13}\) | \(-\frac{5}{12}\) | \(-\frac{12}{5}\) |
| 7 | \(\mathrm{tg}\alpha=-2{,}4,\ \frac{\pi}{2}<\alpha<\pi\) | \(\frac{12}{13}\) | \(-\frac{5}{13}\) | \(-\frac{12}{5}\) | \(-\frac{5}{12}\) |
| 8 | \(\mathrm{ctg}\alpha=\frac{7}{24},\ \pi<\alpha<\frac{3\pi}{2}\) | \(-\frac{24}{25}\) | \(-\frac{7}{25}\) | \(\frac{24}{7}\) | \(\frac{7}{24}\) |
Как рассуждать
Сначала по промежутку определяем четверть и знаки:
- I четверть: все функции положительны;
- II: \(\sin>0,\ \cos<0\), поэтому \(\mathrm{tg}<0,\ \mathrm{ctg}<0\);
- III: \(\sin<0,\ \cos<0\), поэтому \(\mathrm{tg}>0,\ \mathrm{ctg}>0\);
- IV: \(\sin<0,\ \cos>0\), поэтому \(\mathrm{tg}<0,\ \mathrm{ctg}<0\).
Используем основные связи: \[ \sin^2\alpha+\cos^2\alpha=1, \qquad \mathrm{tg}\alpha=\frac{\sin\alpha}{\cos\alpha}, \qquad \mathrm{ctg}\alpha=\frac{1}{\mathrm{tg}\alpha}. \]
Если известен тангенс: \[ 1+\mathrm{tg}^2\alpha=\frac{1}{\cos^2\alpha}. \]
Если известен котангенс: \[ 1+\mathrm{ctg}^2\alpha=\frac{1}{\sin^2\alpha}. \]
Пример 1
Для № 1 дано \(\cos\alpha=\frac{5}{13}\), угол в IV четверти.
\[ \sin^2\alpha=1-\left(\frac{5}{13}\right)^2 =1-\frac{25}{169} =\frac{144}{169}. \]
Значит, \(|\sin\alpha|=\frac{12}{13}\). В IV четверти синус отрицателен, поэтому
\[ \sin\alpha=-\frac{12}{13}. \]
Далее:
\[ \mathrm{tg}\alpha =\frac{\sin\alpha}{\cos\alpha} =\frac{-\frac{12}{13}}{\frac{5}{13}} =-\frac{12}{5}, \qquad \mathrm{ctg}\alpha=-\frac{5}{12}. \]
Пример 3
Для № 3 дано \(\mathrm{tg}\alpha=\frac{15}{8}\), угол в III четверти.
\[ 1+\mathrm{tg}^2\alpha =1+\left(\frac{15}{8}\right)^2 =1+\frac{225}{64} =\frac{289}{64}. \]
Тогда
\[ \cos^2\alpha=\frac{64}{289}, \qquad |\cos\alpha|=\frac{8}{17}. \]
В III четверти косинус отрицателен, значит, \(\cos\alpha=-\frac{8}{17}\). Затем:
\[ \sin\alpha=\mathrm{tg}\alpha\cdot\cos\alpha =\frac{15}{8}\cdot\left(-\frac{8}{17}\right) =-\frac{15}{17}, \qquad \mathrm{ctg}\alpha=\frac{8}{15}. \]
Проверка: \(\sin^2\alpha+\cos^2\alpha=1\), а знаки совпадают с нужной четвертью.
Обсуждение
8 сообщенийБез лишних слов
| № | \(\sin\alpha\) | \(\cos\alpha\) | \(\mathrm{tg}\,\alpha\) | \(\mathrm{ctg}\,\alpha\) |
|---|---|---|---|---|
| 1 | \(-\frac{12}{13}\) | \(\frac{5}{13}\) | \(-\frac{12}{5}\) | \(-\frac{5}{12}\) |
| 2 | \(\frac45\) | \(-\frac35\) | \(-\frac43\) | \(-\frac34\) |
| 3 | \(-\frac{15}{17}\) | \(-\frac{8}{17}\) | \(\frac{15}{8}\) | \(\frac{8}{15}\) |
| 4 | \(-\frac{\sqrt{10}}{10}\) | \(\frac{3\sqrt{10}}{10}\) | \(-\frac13\) | \(-3\) |
| 5 | \(\frac35\) | \(\frac45\) | \(\frac34\) | \(\frac43\) |
| 6 | \(-\frac{5}{13}\) | \(\frac{12}{13}\) | \(-\frac{5}{12}\) | \(-\frac{12}{5}\) |
| 7 | \(\frac{12}{13}\) | \(-\frac{5}{13}\) | \(-\frac{12}{5}\) | \(-\frac{5}{12}\) |
| 8 | \(-\frac{24}{25}\) | \(-\frac{7}{25}\) | \(\frac{24}{7}\) | \(\frac{7}{24}\) |
Без лишних слов
| № | \(\sin\alpha\) | \(\cos\alpha\) | \(\mathrm{tg}\,\alpha\) | \(\mathrm{ctg}\,\alpha\) |
|---|---|---|---|---|
| 1 | \(-\frac{12}{13}\) | \(\frac{5}{13}\) | \(-\frac{12}{5}\) | \(-\frac{5}{12}\) |
| 2 | \(\frac45\) | \(-\frac35\) | \(-\frac43\) | \(-\frac34\) |
| 3 | \(-\frac{15}{17}\) | \(-\frac{8}{17}\) | \(\frac{15}{8}\) | \(\frac{8}{15}\) |
| 4 | \(-\frac{\sqrt{10}}{10}\) | \(\frac{3\sqrt{10}}{10}\) | \(-\frac13\) | \(-3\) |
| 5 | \(\frac35\) | \(\frac45\) | \(\frac34\) | \(\frac43\) |
| 6 | \(-\frac{5}{13}\) | \(\frac{12}{13}\) | \(-\frac{5}{12}\) | \(-\frac{12}{5}\) |
| 7 | \(\frac{12}{13}\) | \(-\frac{5}{13}\) | \(-\frac{12}{5}\) | \(-\frac{5}{12}\) |
| 8 | \(-\frac{24}{25}\) | \(-\frac{7}{25}\) | \(\frac{24}{7}\) | \(\frac{7}{24}\) |
С решением
| № | Решение | Ответ: \(\sin\alpha,\ \cos\alpha,\ \mathrm{tg}\,\alpha,\ \mathrm{ctg}\,\alpha\) |
|---|---|---|
| 1 | \(\sin^2=1-\frac{25}{169}=\frac{144}{169}\); IV ч. \(\Rightarrow \sin<0\) | \(-\frac{12}{13},\ \frac{5}{13},\ -\frac{12}{5},\ -\frac{5}{12}\) |
| 2 | \(\cos^2=1-0{,}64=0{,}36\); II ч. \(\Rightarrow \cos<0\) | \(\frac45,\ -\frac35,\ -\frac43,\ -\frac34\) |
| 3 | \(1+\mathrm{tg}^2=1+\frac{225}{64}=\frac{289}{64}\); III ч. \(\Rightarrow \cos<0\) | \(-\frac{15}{17},\ -\frac{8}{17},\ \frac{15}{8},\ \frac{8}{15}\) |
| 4 | \(1+\mathrm{ctg}^2=1+9=10\); IV ч. \(\Rightarrow \sin<0\) | \(-\frac{\sqrt{10}}{10},\ \frac{3\sqrt{10}}{10},\ -\frac13,\ -3\) |
| 5 | \(\sin^2=1-0{,}64=0{,}36\); I ч. \(\Rightarrow \sin>0\) | \(\frac35,\ \frac45,\ \frac34,\ \frac43\) |
| 6 | \(\cos^2=1-\frac{25}{169}=\frac{144}{169}\); IV ч. \(\Rightarrow \cos>0\) | \(-\frac{5}{13},\ \frac{12}{13},\ -\frac{5}{12},\ -\frac{12}{5}\) |
| 7 | \(1+\mathrm{tg}^2=1+5{,}76=6{,}76=\frac{169}{25}\); II ч. \(\Rightarrow \cos<0\), \(\sin>0\) | \(\frac{12}{13},\ -\frac{5}{13},\ -\frac{12}{5},\ -\frac{5}{12}\) |
| 8 | \(1+\mathrm{ctg}^2=1+\frac{49}{576}=\frac{625}{576}\); III ч. \(\Rightarrow \sin<0\), \(\cos<0\) | \(-\frac{24}{25},\ -\frac{7}{25},\ \frac{24}{7},\ \frac{7}{24}\) |
485 реши
485. Найти значение выражения
\(\sin 73^\circ\cos 17^\circ+\cos 73^\circ\sin 17^\circ=\sin(73^\circ+17^\circ)=\sin 90^\circ=1\)
\(\sin 73^\circ\cos 13^\circ-\cos 73^\circ\sin 13^\circ=\sin(73^\circ-13^\circ)=\sin 60^\circ=\frac{\sqrt3}{2}\)
\(\sin\frac{5\pi}{12}\cos\frac{\pi}{12}+\sin\frac{\pi}{12}\cos\frac{5\pi}{12}=\sin\left(\frac{5\pi}{12}+\frac{\pi}{12}\right)=\sin\frac{\pi}{2}=1\)
\(\sin\frac{7\pi}{12}\cos\frac{\pi}{12}-\sin\frac{\pi}{12}\cos\frac{7\pi}{12}=\sin\left(\frac{7\pi}{12}-\frac{\pi}{12}\right)=\sin\frac{\pi}{2}=1\)
Ответ: 1) \(1\); 2) \(\frac{\sqrt3}{2}\); 3) \(1\); 4) \(1\).
Использованы формулы: \[ \sin(\alpha+\beta)=\sin\alpha\cos\beta+\cos\alpha\sin\beta, \] \[ \sin(\alpha-\beta)=\sin\alpha\cos\beta-\cos\alpha\sin\beta. \]
