ГДЗ по Алгебре 10 Класс Номер 37.6 Профильный Уровень Мордкович — Подробные Ответы

а) $y_n = \sin \frac{n\pi}{2} — \operatorname{ctg} \frac{\pi}{4}(2n + 1)$;

$y_1 = \sin \frac{\pi}{2} — \operatorname{ctg} \frac{\pi(2+1)}{4} = \sin \frac{\pi}{2} — \operatorname{ctg} \frac{3\pi}{4} = 1 + 1 = 2;$

$y_2 = \sin \frac{2\pi}{2} — \operatorname{ctg} \frac{\pi(2 \cdot 2 + 1)}{4} = \sin \pi — \operatorname{ctg} \frac{5\pi}{4} = 0 — 1 = -1;$

$y_3 = \sin \frac{3\pi}{2} — \operatorname{ctg} \frac{\pi(2 \cdot 3 + 1)}{4} = \sin \frac{3\pi}{2} — \operatorname{ctg} \frac{7\pi}{4} = -1 + 1 = 0;$

$y_4 = \sin \frac{4\pi}{2} — \operatorname{ctg} \frac{\pi(2 \cdot 4 + 1)}{4} = \sin 2\pi — \operatorname{ctg} \frac{9\pi}{4} = -\operatorname{ctg} \frac{\pi}{4} = -1;$

$y_5 = \sin \frac{5\pi}{2} — \operatorname{ctg} \frac{\pi(2 \cdot 5 + 1)}{4} = \sin \frac{\pi}{2} — \operatorname{ctg} \frac{11\pi}{4} = \sin \frac{\pi}{2} — \operatorname{ctg} \frac{3\pi}{4} = 2;$

б) $y_n = \cos \frac{n\pi}{2} + \operatorname{tg} \frac{\pi}{4} \cdot (2n + 1)$;

$y_1 = \cos \frac{\pi}{2} + \operatorname{tg} \frac{\pi(2+1)}{4} = \cos \frac{\pi}{2} + \operatorname{tg} \frac{3\pi}{4} = 0 — 1 = -1;$

$y_2 = \cos \frac{2\pi}{2} + \operatorname{tg} \frac{\pi(2 \cdot 2 + 1)}{4} = \cos \pi + \operatorname{tg} \frac{5\pi}{4} = -1 + 1 = 0;$

$y_3 = \cos \frac{3\pi}{2} + \operatorname{tg} \frac{\pi(2 \cdot 3 + 1)}{4} = \cos \frac{\pi}{2} + \operatorname{tg} \frac{7\pi}{4} = 0 — 1 = -1;$

$y_4 = \cos \frac{4\pi}{2} + \operatorname{tg} \frac{\pi(2 \cdot 4 + 1)}{4} = \cos 2\pi + \operatorname{tg} \frac{9\pi}{4} = 1 + \operatorname{tg} \frac{\pi}{4} = 1 + 1 = 2;$

$y_5 = \cos \frac{5\pi}{2} + \operatorname{tg} \frac{\pi(2 \cdot 5 + 1)}{4} = \cos \frac{\pi}{2} + \operatorname{tg} \frac{11\pi}{4} = \cos \frac{\pi}{2} + \operatorname{tg} \frac{3\pi}{4} = -1;$

в) $y_n = n \sin \frac{n\pi}{2} + n^2 \cos \frac{n\pi}{2}$;

$y_1 = \sin \frac{\pi}{2} + \cos \frac{\pi}{2} = 1 + 0 = 1;$

$y_2 = 2 \sin \frac{2\pi}{2} + 2^2 \cos \frac{2\pi}{2} = 2 \sin \pi + 4 \cos \pi = 2 \cdot 0 + 4 \cdot (-1) = -4;$

$y_3 = 3 \sin \frac{3\pi}{2} + 3^2 \cos \frac{3\pi}{2} = 3 \cdot (-1) + 9 \cdot 0 = -3;$

$y_4 = 4 \sin \frac{4\pi}{2} + 4^2 \cos \frac{4\pi}{2} = 4 \sin 2\pi + 16 \cos 2\pi = 4 \cdot 0 + 16 \cdot 1 = 16;$

$y_5 = 5 \sin \frac{5\pi}{2} + 5^2 \cos \frac{5\pi}{2} = 5 \sin \frac{\pi}{2} + 25 \cos \frac{\pi}{2} = 5 \cdot 1 + 25 \cdot 0 = 5;$

г) $y_n = \sin \frac{n\pi}{4} — n \cos \frac{n\pi}{4}$;

$y_1 = \sin \frac{\pi}{4} — \cos \frac{\pi}{4} = \frac{\sqrt{2}}{2} — \frac{\sqrt{2}}{2} = 0;$

$y_2 = \sin \frac{2\pi}{4} — 2 \cos \frac{2\pi}{4} = \sin \frac{\pi}{2} — 2 \cos \frac{\pi}{2} = 1 — 2 \cdot 0 = 1;$

$y_3 = \sin \frac{3\pi}{4} — 3 \cos \frac{3\pi}{4} = \frac{\sqrt{2}}{2} — 3 \left( -\frac{\sqrt{2}}{2} \right) = \frac{\sqrt{2}}{2} + \frac{3\sqrt{2}}{2} = \frac{4\sqrt{2}}{2} = 2\sqrt{2};$

$y_4 = \sin \frac{4\pi}{4} — 4 \cos \frac{4\pi}{4} = \sin \pi — 4 \cos \pi = 0 — 4 \cdot (-1) = 4;$

$y_5 = \sin \frac{5\pi}{4} — 5 \cos \frac{5\pi}{4} = -\frac{\sqrt{2}}{2} — 5 \left( -\frac{\sqrt{2}}{2} \right) = -\frac{\sqrt{2}}{2} + \frac{5\sqrt{2}}{2} = \frac{4\sqrt{2}}{2} = 2\sqrt{2};$

Рассмотрим последовательности $y_n$, заданные следующими формулами:

а) $y_n = \sin \frac{n\pi}{2} — \operatorname{ctg} \left( \frac{\pi}{4} (2n + 1) \right)$

б) $y_n = \cos \frac{n\pi}{2} + \operatorname{tg} \left( \frac{\pi}{4} (2n + 1) \right)$

в) $y_n = n \sin \frac{n\pi}{2} + n^2 \cos \frac{n\pi}{2}$

г) $y_n = \sin \frac{n\pi}{4} — n \cos \frac{n\pi}{4}$

Необходимо найти первые 5 значений для каждой последовательности $y_n$ при $n = 1, 2, 3, 4, 5$.

Часть а) $y_n = \sin \frac{n\pi}{2} — \operatorname{ctg} \left( \frac{\pi}{4}(2n + 1) \right)$

Рассмотрим выражение для каждого $n$:

1. Для $n = 1$:

$$y_1 = \sin \frac{1\pi}{2} — \operatorname{ctg} \left( \frac{\pi}{4}(2 \cdot 1 + 1) \right)$$ $$= \sin \frac{\pi}{2} — \operatorname{ctg} \left( \frac{\pi}{4} \cdot 3 \right)$$ $$= 1 — \operatorname{ctg} \frac{3\pi}{4}$$

Так как $\operatorname{ctg} \theta = \frac{1}{\tan \theta}$, и $\tan \frac{3\pi}{4} = -1$, то:

$$\operatorname{ctg} \frac{3\pi}{4} = -1$$

Следовательно:

$$y_1 = 1 — (-1) = 1 + 1 = 2$$

2. Для $n = 2$:

$$y_2 = \sin \frac{2\pi}{2} — \operatorname{ctg} \left( \frac{\pi}{4}(2 \cdot 2 + 1) \right)$$ $$= \sin \pi — \operatorname{ctg} \left( \frac{\pi}{4} \cdot 5 \right)$$ $$= 0 — \operatorname{ctg} \frac{5\pi}{4}$$

Так как $\operatorname{ctg} \theta = \frac{1}{\tan \theta}$, и $\tan \frac{5\pi}{4} = 1$, то:

$$\operatorname{ctg} \frac{5\pi}{4} = 1$$

Следовательно:

$$y_2 = 0 — 1 = -1$$

3. Для $n = 3$:

$$y_3 = \sin \frac{3\pi}{2} — \operatorname{ctg} \left( \frac{\pi}{4}(2 \cdot 3 + 1) \right)$$ $$= \sin \frac{3\pi}{2} — \operatorname{ctg} \left( \frac{\pi}{4} \cdot 7 \right)$$ $$= -1 — \operatorname{ctg} \frac{7\pi}{4}$$

Так как $\operatorname{ctg} \theta = \frac{1}{\tan \theta}$, и $\tan \frac{7\pi}{4} = -1$, то:

$$\operatorname{ctg} \frac{7\pi}{4} = -1$$

Следовательно:

$$y_3 = -1 — (-1) = -1 + 1 = 0$$

4. Для $n = 4$:

$$y_4 = \sin \frac{4\pi}{2} — \operatorname{ctg} \left( \frac{\pi}{4}(2 \cdot 4 + 1) \right)$$ $$= \sin 2\pi — \operatorname{ctg} \left( \frac{\pi}{4} \cdot 9 \right)$$ $$= 0 — \operatorname{ctg} \frac{9\pi}{4}$$

Так как $\operatorname{ctg} \theta = \frac{1}{\tan \theta}$, и $\tan \frac{9\pi}{4} = 1$, то:

$$\operatorname{ctg} \frac{9\pi}{4} = 1$$

Следовательно:

$$y_4 = 0 — 1 = -1$$

5. Для $n = 5$:

$$y_5 = \sin \frac{5\pi}{2} — \operatorname{ctg} \left( \frac{\pi}{4}(2 \cdot 5 + 1) \right)$$ $$= \sin \frac{\pi}{2} — \operatorname{ctg} \left( \frac{\pi}{4} \cdot 11 \right)$$ $$= 1 — \operatorname{ctg} \frac{11\pi}{4}$$

Так как $\operatorname{ctg} \theta = \frac{1}{\tan \theta}$, и $\tan \frac{11\pi}{4} = -1$, то:

$$\operatorname{ctg} \frac{11\pi}{4} = -1$$

Следовательно:

$$y_5 = 1 — (-1) = 1 + 1 = 2$$

Часть б) $y_n = \cos \frac{n\pi}{2} + \operatorname{tg} \left( \frac{\pi}{4}(2n + 1) \right)$

1. Для $n = 1$:

$$y_1 = \cos \frac{\pi}{2} + \operatorname{tg} \left( \frac{\pi}{4}(2 \cdot 1 + 1) \right)$$ $$= \cos \frac{\pi}{2} + \operatorname{tg} \frac{3\pi}{4}$$ $$= 0 + (-1) = -1$$

2. Для $n = 2$:

$$y_2 = \cos \frac{2\pi}{2} + \operatorname{tg} \left( \frac{\pi}{4}(2 \cdot 2 + 1) \right)$$ $$= \cos \pi + \operatorname{tg} \frac{5\pi}{4}$$ $$= -1 + 1 = 0$$

3. Для $n = 3$:

$$y_3 = \cos \frac{3\pi}{2} + \operatorname{tg} \left( \frac{\pi}{4}(2 \cdot 3 + 1) \right)$$ $$= \cos \frac{3\pi}{2} + \operatorname{tg} \frac{7\pi}{4}$$ $$= 0 + (-1) = -1$$

4. Для $n = 4$:

$$y_4 = \cos \frac{4\pi}{2} + \operatorname{tg} \left( \frac{\pi}{4}(2 \cdot 4 + 1) \right)$$ $$= \cos 2\pi + \operatorname{tg} \frac{9\pi}{4}$$ $$= 1 + 1 = 2$$

5. Для $n = 5$:

$$y_5 = \cos \frac{5\pi}{2} + \operatorname{tg} \left( \frac{\pi}{4}(2 \cdot 5 + 1) \right)$$ $$= \cos \frac{\pi}{2} + \operatorname{tg} \frac{11\pi}{4}$$ $$= 0 + (-1) = -1$$

Часть в) $y_n = n \sin \frac{n\pi}{2} + n^2 \cos \frac{n\pi}{2}$

1. Для $n = 1$:

$$y_1 = 1 \cdot \sin \frac{\pi}{2} + 1^2 \cdot \cos \frac{\pi}{2}$$ $$= 1 \cdot 1 + 1^2 \cdot 0 = 1 + 0 = 1$$

2. Для $n = 2$:

$$y_2 = 2 \cdot \sin \pi + 2^2 \cdot \cos \pi$$ $$= 2 \cdot 0 + 4 \cdot (-1) = 0 — 4 = -4$$

3. Для $n = 3$:

$$y_3 = 3 \cdot \sin \frac{3\pi}{2} + 3^2 \cdot \cos \frac{3\pi}{2}$$ $$= 3 \cdot (-1) + 9 \cdot 0 = -3 + 0 = -3$$

4. Для $n = 4$:

$$y_4 = 4 \cdot \sin 2\pi + 4^2 \cdot \cos 2\pi$$ $$= 4 \cdot 0 + 16 \cdot 1 = 0 + 16 = 16$$

5. Для $n = 5$:

$$y_5 = 5 \cdot \sin \frac{5\pi}{2} + 5^2 \cdot \cos \frac{5\pi}{2}$$ $$= 5 \cdot 1 + 25 \cdot 0 = 5 + 0 = 5$$

Часть г) $y_n = \sin \frac{n\pi}{4} — n \cos \frac{n\pi}{4}$

1. Для $n = 1$:

$$y_1 = \sin \frac{\pi}{4} — 1 \cdot \cos \frac{\pi}{4}$$ $$= \frac{\sqrt{2}}{2} — \frac{\sqrt{2}}{2} = 0$$

2. Для $n = 2$:

$$y_2 = \sin \frac{2\pi}{4} — 2 \cdot \cos \frac{2\pi}{4}$$ $$= \sin \frac{\pi}{2} — 2 \cdot \cos \frac{\pi}{2}$$ $$= 1 — 2 \cdot 0 = 1$$

3. Для $n = 3$:

$$y_3 = \sin \frac{3\pi}{4} — 3 \cdot \cos \frac{3\pi}{4}$$ $$= \frac{\sqrt{2}}{2} — 3 \left( -\frac{\sqrt{2}}{2} \right)$$ $$= \frac{\sqrt{2}}{2} + \frac{3\sqrt{2}}{2} = \frac{4\sqrt{2}}{2} = 2\sqrt{2}$$

4. Для $n = 4$:

$$y_4 = \sin \pi — 4 \cdot \cos \pi$$ $$= 0 — 4 \cdot (-1) = 4$$

5. Для $n = 5$:

$$y_5 = \sin \frac{5\pi}{4} — 5 \cdot \cos \frac{5\pi}{4}$$ $$= -\frac{\sqrt{2}}{2} — 5 \left( -\frac{\sqrt{2}}{2} \right)$$ $$= -\frac{\sqrt{2}}{2} + \frac{5\sqrt{2}}{2} = \frac{4\sqrt{2}}{2} = 2\sqrt{2}$$

Ответы:

а) $y_1 = 2, y_2 = -1, y_3 = 0, y_4 = -1, y_5 = 2$

б) $y_1 = -1, y_2 = 0, y_3 = -1, y_4 = 2, y_5 = -1$

в) $y_1 = 1, y_2 = -4, y_3 = -3, y_4 = 16, y_5 = 5$

г) $y_1 = 0, y_2 = 1, y_3 = 2\sqrt{2}, y_4 = 4, y_5 = 2\sqrt{2}$