Упр.8.22 ГДЗ Никольский 10 класс (Алгебра)

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1. а) $$0<\alpha<\frac{\pi}{2},\quad \cos\alpha=\frac{3}{5}$$

   $$\sin^2\alpha=1-\cos^2\alpha=1-\frac{9}{25}=\frac{16}{25}$$
   $$\sin\alpha=\frac{4}{5}$$
   $$\tg\alpha=\frac{\sin\alpha}{\cos\alpha}=\frac{4/5}{3/5}=\frac{4}{3}$$
   $$\ctg\alpha=\frac{1}{\tg\alpha}=\frac{3}{4}$$

2. б) $$\frac{\pi}{2}<\alpha<\pi,\quad \sin\alpha=\frac{1}{2}$$

   $$\cos^2\alpha=1-\sin^2\alpha=1-\frac{1}{4}=\frac{3}{4}$$
   $$\cos\alpha=-\frac{\sqrt{3}}{2}$$
   $$\tg\alpha=\frac{\sin\alpha}{\cos\alpha}=\frac{1/2}{-\sqrt{3}/2}=-\frac{1}{\sqrt{3}}=-\frac{\sqrt{3}}{3}$$
   $$\ctg\alpha=\frac{1}{\tg\alpha}=-\sqrt{3}$$

3. в) $$\pi<\alpha<\frac{3\pi}{2},\quad \cos\alpha=-0{,}6$$

   $$\sin^2\alpha=1-\cos^2\alpha=1-0{,}36=0{,}64$$
   $$\sin\alpha=-0{,}8$$
   $$\tg\alpha=\frac{\sin\alpha}{\cos\alpha}=\frac{-0{,}8}{-0{,}6}=\frac{4}{3}$$
   $$\ctg\alpha=\frac{1}{\tg\alpha}=\frac{3}{4}$$

4. г) $$\frac{3\pi}{2}<\alpha<2\pi,\quad \sin\alpha=-0{,}8$$

   $$\cos^2\alpha=1-\sin^2\alpha=1-0{,}64=0{,}36$$
   $$\cos\alpha=0{,}6$$
   $$\tg\alpha=\frac{\sin\alpha}{\cos\alpha}=\frac{-0{,}8}{0{,}6}=-\frac{4}{3}$$
   $$\ctg\alpha=\frac{1}{\tg\alpha}=-\frac{3}{4}$$

5. д) $$0<\alpha<\frac{\pi}{2},\quad \tg\alpha=2{,}4$$

   $$\tg\alpha=\frac{12}{5}$$
   $$1+\tg^2\alpha=\frac{1}{\cos^2\alpha}$$
   $$1+\frac{144}{25}=\frac{169}{25}=\frac{1}{\cos^2\alpha}$$
   $$\cos^2\alpha=\frac{25}{169}$$
   $$\cos\alpha=\frac{5}{13}$$
   $$\sin\alpha=\sqrt{1-\cos^2\alpha}=\sqrt{1-\frac{25}{169}}=\frac{12}{13}$$
   $$\ctg\alpha=\frac{1}{\tg\alpha}=\frac{5}{12}$$

6. е) $$\frac{\pi}{2}<\alpha<\pi,\quad \tg\alpha=-1$$

   $$1+\tg^2\alpha=\frac{1}{\cos^2\alpha}$$
   $$1+1=2=\frac{1}{\cos^2\alpha}$$
   $$\cos^2\alpha=\frac{1}{2}$$
   $$\cos\alpha=-\frac{\sqrt{2}}{2}$$
   $$\sin\alpha=\sqrt{1-\cos^2\alpha}=\sqrt{1-\frac{1}{2}}=\frac{\sqrt{2}}{2}$$
   $$\ctg\alpha=\frac{1}{\tg\alpha}=-1$$

7. ж) $$-\frac{\pi}{2}<\alpha<0,\quad \tg\alpha=-\frac{5}{12}$$

   $$\ctg\alpha=\frac{1}{\tg\alpha}=-\frac{12}{5}$$
   $$1+\tg^2\alpha=\frac{1}{\cos^2\alpha}$$
   $$1+\frac{25}{144}=\frac{169}{144}=\frac{1}{\cos^2\alpha}$$
   $$\cos^2\alpha=\frac{144}{169}$$
   $$\cos\alpha=\frac{12}{13}$$
   $$\sin\alpha=\tg\alpha\cdot\cos\alpha=-\frac{5}{12}\cdot\frac{12}{13}=-\frac{5}{13}$$

8. з) $$\pi<\alpha<\frac{3\pi}{2},\quad \ctg\alpha=1$$

   $$\tg\alpha=\frac{1}{\ctg\alpha}=1$$
   $$1+\tg^2\alpha=\frac{1}{\cos^2\alpha}$$
   $$1+1=2=\frac{1}{\cos^2\alpha}$$
   $$\cos^2\alpha=\frac{1}{2}$$
   $$\cos\alpha=-\frac{\sqrt{2}}{2}$$
   $$\sin\alpha=\tg\alpha\cdot\cos\alpha=1\cdot\left(-\frac{\sqrt{2}}{2}\right)=-\frac{\sqrt{2}}{2}$$

Ответ

а) $$\sin\alpha=\frac{4}{5},\ \tg\alpha=\frac{4}{3},\ \ctg\alpha=\frac{3}{4}$$
б) $$\cos\alpha=-\frac{\sqrt{3}}{2},\ \tg\alpha=-\frac{\sqrt{3}}{3},\ \ctg\alpha=-\sqrt{3}$$
в) $$\sin\alpha=-0{,}8,\ \tg\alpha=\frac{4}{3},\ \ctg\alpha=\frac{3}{4}$$
г) $$\cos\alpha=0{,}6,\ \tg\alpha=-\frac{4}{3},\ \ctg\alpha=-\frac{3}{4}$$
д) $$\sin\alpha=\frac{12}{13},\ \cos\alpha=\frac{5}{13},\ \ctg\alpha=\frac{5}{12}$$
е) $$\sin\alpha=\frac{\sqrt{2}}{2},\ \cos\alpha=-\frac{\sqrt{2}}{2},\ \ctg\alpha=-1$$
ж) $$\sin\alpha=-\frac{5}{13},\ \cos\alpha=\frac{12}{13},\ \ctg\alpha=-\frac{12}{5}$$
з) $$\sin\alpha=-\frac{\sqrt{2}}{2},\ \cos\alpha=-\frac{\sqrt{2}}{2},\ \tg\alpha=1$$