Задачи по тригонометрия за ЗИП - 2 част

Задачите са предоставени от:
г-н Николай Чакъров от Шумен
преподавател в ПГОХХТ "Проф. д-р Асен Златаров"

Математическа гимназия - Варна

Зад. 1 Докажете тъждествата:

А) $\frac{1}{2}(\cos\alpha + \sqrt{3}\sin\alpha) = \sin(30^\circ + \alpha)$

Решение

$\frac{1}{2}\cos\alpha + \frac{\sqrt{3}}{2}\sin\alpha = \sin 30^\circ \cos\alpha + \cos 30^\circ \sin\alpha = \sin(30^\circ + \alpha)$

Б) $\cos\alpha - \sqrt{3}\sin\alpha = 2\cos(60^\circ + \alpha)$

Решение

Разделяме двете страни на $2$:

$\frac{1}{2}\cos\alpha - \frac{\sqrt{3}}{2}\sin\alpha = \cos 60^\circ \cos\alpha - \sin 60^\circ \sin\alpha = \cos(60^\circ + \alpha)$

В) $\operatorname{tg} 20^\circ + \operatorname{tg} 25^\circ + \operatorname{tg} 20^\circ \operatorname{tg} 25^\circ = 1$

Решение

$\operatorname{tg} 20^\circ + \operatorname{tg} 25^\circ = 1 - \operatorname{tg} 20^\circ \operatorname{tg} 25^\circ \;\Leftrightarrow\; \frac{\operatorname{tg} 20^\circ + \operatorname{tg} 25^\circ}{1 - \operatorname{tg} 20^\circ \operatorname{tg} 25^\circ} = 1$

$\Leftrightarrow\; \operatorname{tg}(20^\circ + 25^\circ) = \operatorname{tg} 45^\circ = 1$

Г) $\operatorname{tg} 20^\circ + \operatorname{tg} 40^\circ + \sqrt{3}\operatorname{tg} 20^\circ \operatorname{tg} 40^\circ = \sqrt{3}$

Решение

$\operatorname{tg} 20^\circ + \operatorname{tg} 40^\circ = \sqrt{3}\,(1 - \operatorname{tg} 20^\circ \operatorname{tg} 40^\circ) \;\Leftrightarrow\; \frac{\operatorname{tg} 20^\circ + \operatorname{tg} 40^\circ}{1 - \operatorname{tg} 20^\circ \operatorname{tg} 40^\circ} = \sqrt{3}$

$\Leftrightarrow\; \operatorname{tg}(20^\circ + 40^\circ) = \operatorname{tg} 60^\circ = \sqrt{3}$

Д) $\sin(\alpha + \beta)\sin(\alpha - \beta) = \sin^2\alpha - \sin^2\beta$

Е) $\cos(\alpha + \beta)\cos(\alpha - \beta) = \cos^2\alpha - \sin^2\beta$

Ж) $\sin^2\left(\alpha - \frac{\pi}{6}\right) + \sin^2\left(\alpha + \frac{\pi}{6}\right) - \sin^2\alpha = \frac{1}{2}$

Зад. 2 В триъгълника $ABC$ $\cos A = \frac{12}{13}$, $\cos B = \frac{7}{25}$. Да се пресметне синусът на външния ъгъл на триъгълника, несъседен на тези два ъгъла.

Решение

Външният ъгъл при върха $C$ е $180^\circ - C$, а $C = 180^\circ - (A + B)$:

$\sin(180^\circ - C) = \sin C = \sin(180^\circ - (A + B)) = \sin(A + B) = \sin A \cos B + \cos A \sin B$

Ъглите $A$ и $B$ са между $0^\circ$ и $180^\circ$, затова синусите им са положителни:

$\cos A = \frac{12}{13} \;\Rightarrow\; \sin A = \sqrt{1 - \left(\frac{12}{13}\right)^2} = \sqrt{\frac{25}{169}} = \frac{5}{13}$

$\cos B = \frac{7}{25} \;\Rightarrow\; \sin B = \sqrt{1 - \left(\frac{7}{25}\right)^2} = \sqrt{\frac{576}{625}} = \frac{24}{25}$

$\sin(180^\circ - C) = \frac{5}{13} \cdot \frac{7}{25} + \frac{12}{13} \cdot \frac{24}{25} = \frac{35 + 288}{325} = \frac{323}{325}$

Зад. 3 Докажете, че ако ъглите на един триъгълник удовлетворяват равенството $\sin\alpha = 2\sin\beta\cos\gamma$, то той е равнобедрен.

Решение

От $\gamma = 180^\circ - (\alpha + \beta)$ следва $\cos\gamma = \cos(180^\circ - (\alpha + \beta)) = -\cos(\alpha + \beta)$, тогава

$\sin\alpha = -2\sin\beta\cos(\alpha + \beta)$

$\sin\alpha + 2\sin\beta\,(\cos\alpha\cos\beta - \sin\alpha\sin\beta) = 0$

$\sin\alpha + 2\sin\beta\cos\beta\cos\alpha - 2\sin^2\beta\sin\alpha = 0$

$\sin\alpha\,(1 - 2\sin^2\beta) + \sin 2\beta\cos\alpha = 0 \qquad (1)$

Използваме формулата за понижаване на степента $\sin^2\beta = \frac{1 - \cos 2\beta}{2}$:

$1 - 2\sin^2\beta = 1 - (1 - \cos 2\beta) = \cos 2\beta$

$(1) \;\Leftrightarrow\; \sin\alpha\cos 2\beta + \sin 2\beta\cos\alpha = 0 \;\Leftrightarrow\; \sin(\alpha + 2\beta) = 0 \;\Leftrightarrow\; \alpha + 2\beta = k\pi, \; k \in \mathbb{Z}$

Понеже $\alpha + \beta \lt \pi$ и $\beta \lt \pi$, то $0 \lt \alpha + 2\beta \lt 2\pi$, затова $k = 1$ и $\alpha = \pi - 2\beta$. Тогава $\gamma = \pi - \alpha - \beta = \beta$, т.е. триъгълникът е равнобедрен.

Зад. 4

А) Представете като произведение:

1. $1 + \cos 3\alpha$
2. $1 - \cos\frac{\alpha}{2}$
3. $\operatorname{tg}\alpha - \sin\alpha$
4. $\cos 3\alpha - \sin\alpha$
5. $\operatorname{tg}\alpha + \operatorname{tg}\beta$
6. $\operatorname{cotg}\alpha - \operatorname{cotg}\beta$
7. $\operatorname{tg} 2\alpha - \operatorname{cotg}\alpha$
8. $1 + \operatorname{tg}\alpha$
9. $\operatorname{cotg}\alpha - 1$
10. $2\operatorname{cotg} 2\alpha - 1$
11. $\sqrt{3} - 2\cos\alpha$
12. $\sqrt{2} + 2\sin\alpha$
13. $\sqrt{3} + \operatorname{tg}\alpha$
14. $1 - \sqrt{3}\operatorname{cotg} 4\alpha$
15. $\sqrt{3} + 4\cos 3\alpha \sin 3\alpha$
16. $\frac{1}{2\cos\alpha} - \operatorname{tg}\alpha$
17. $\operatorname{cotg}\frac{\alpha}{2} + \frac{1}{\sqrt{2}\sin\frac{\alpha}{2}}$
18. $\operatorname{tg}\alpha + \frac{\sin\beta}{2\cos^2\alpha}$

Б) Представете като произведение:

1. $1 - \frac{\sin^2\alpha}{\sin^2\beta}$
2. $\frac{1}{2\cos\frac{\alpha}{2}} - 1$
3. $\frac{1}{1 - 4\sin^2\frac{\alpha}{2}}$
4. $\frac{1}{1 - \operatorname{tg}\alpha} - \frac{1}{1 + \operatorname{tg}\alpha}$
5. $\frac{1}{4\cos\alpha} - \sin^2\alpha$
6. $4\sin^4\frac{\alpha}{2} + \sin^2\frac{\beta}{2} - 4\sin^2\frac{\alpha}{2}$
7. $1 - \operatorname{tg}^2\alpha \operatorname{cotg}\beta$
8. $3\operatorname{cotg}^2\frac{\alpha}{2} - 1$
9. $3\sin^2\alpha - \cos^2\alpha$
10. $1 + \sin\alpha - \cos\alpha$
11. $1 + \operatorname{tg} 2\alpha + \frac{1}{\cos 2\alpha}$
12. $\cos^2\beta - \sin^2\beta \operatorname{cotg}^2\alpha$
13. $1 + \operatorname{cotg}\alpha - \frac{1}{\sin\alpha}$
14. $1 - \sqrt{3}\sin\frac{\alpha}{2} - \cos\alpha$
15. $\sin\alpha + \sin 2\alpha + \sin 3\alpha$
16. $\cos 5\alpha + \sin 6\alpha - \cos 7\alpha$
17. $\cos 8\alpha \cos 10\alpha - \cos\alpha \cos 3\alpha$
18. $\sin 10\alpha \cos 4\alpha - \sin 6\alpha \cos 8\alpha$
19. $\sin(\alpha + \beta)\sin(\alpha - \beta)$
20. $\sin\alpha - 2\sin\left(\frac{\alpha}{2} - \frac{\pi}{4}\right)\cos\left(\frac{\alpha}{2} - \frac{\pi}{4}\right)$

В) Опростете изразите:

1. $\cos(500^\circ + \alpha) + \cos(400^\circ - \alpha)$
2. $\sin(\alpha + \beta) + \cos(\alpha - \beta)$
3. $\sin 2\alpha - 2\sin\alpha$
4. $\frac{1 - \sin 2\alpha}{1 + \sin 2\alpha}$
5. $\frac{\operatorname{tg} 2\alpha - \operatorname{tg}\alpha}{\operatorname{tg} 2\alpha + \operatorname{tg}\alpha}$
6. $\operatorname{cotg}\frac{\alpha}{2} - 2\operatorname{cotg}\alpha$
7. $1 + \operatorname{tg}\alpha \operatorname{tg}\frac{\alpha}{2}$
8. $\frac{1 - \operatorname{tg}\alpha \operatorname{tg}\beta}{1 + \operatorname{tg}\alpha \operatorname{tg}\beta}$
9. $\frac{\sin\alpha + \sin\beta}{\sin\alpha - \sin\beta} \cdot \operatorname{tg}\frac{\alpha - \beta}{2}$
10. $\sin\alpha\cos\alpha + \sin\beta\cos\beta$
11. $\frac{\sin^2\alpha - \sin^2\beta}{\sin\frac{\alpha}{2}\cos\frac{\alpha}{2} - \sin\frac{\beta}{2}\cos\frac{\beta}{2}}$
12. $\frac{\cos\alpha\cos\beta + \sin\alpha\sin\beta}{\cos\alpha - \cos\beta}$
13. $\cos 24^\circ + \cos 48^\circ + \cos 60^\circ + \cos 72^\circ$
14. $\operatorname{tg} 30^\circ + \operatorname{tg} 40^\circ + \operatorname{tg} 50^\circ + \operatorname{tg} 60^\circ$

Г) Проверете тъждествата:

1. $2\sin\alpha \sin 2\alpha + \cos 3\alpha = \cos\alpha$
2. $\frac{1}{2\sin 10^\circ} - 2\sin 70^\circ = 1$
3. $\sin 3\alpha = 4\sin\alpha \sin(60^\circ - \alpha)\sin(60^\circ + \alpha)$
4. $\cos 3\alpha = 4\cos\alpha \cos(60^\circ - \alpha)\cos(60^\circ + \alpha)$

Д) Изчислете:

1. $\sin\frac{\alpha + \beta}{2} \cos\frac{\alpha + \beta}{2}$, ако $\sin\alpha = \frac{2}{3}$, $\sin\beta = \frac{1}{3}$
2. $\sin\frac{\alpha + \beta}{2} \cos\frac{\alpha - \beta}{2}$, ако $\sin\alpha = \frac{2}{3}$, $\sin\beta = \frac{1}{3}$
3. $\cos\frac{\alpha + \beta}{2} \cos\frac{\alpha - \beta}{2}$, ако $\cos\alpha = \frac{2}{3}$, $\sin\beta = \frac{4}{5}$, $\beta \in$ II квадрант
4. $\sin(\alpha + \beta)$ и $\cos(\alpha + \beta)$, ако $\sin\alpha + \sin\beta = a$, $\cos\alpha + \cos\beta = b$

Е) Определете вида на триъгълника според ъглите му, ако:

1. $\sin\alpha = \cos\beta + \cos\gamma$
2. $\sin\alpha = \frac{\sin\beta + \sin\gamma}{\cos\beta + \cos\gamma}$
3. $\frac{\sin\alpha + \sin\beta}{\sin\gamma} = \operatorname{cotg}\frac{\gamma}{2}$

Зад. 5 Пресметнете:

1. $\frac{\sin^2(\alpha + \beta) - \sin^2\alpha - \sin^2\beta}{\sin^2(\alpha + \beta) - \cos^2\alpha - \cos^2\beta}$
2. $\sin^2(\alpha - 2\beta) - \cos^2\alpha - \cos^2 2\beta$
3. $\sin^2(\alpha - 2\beta) - \sin^2 2\alpha - \cos^2\beta$
4. $2\operatorname{cotg}\frac{5\pi + \alpha}{4}\left(1 + \cos\frac{\alpha - \pi}{2}\right) \cdot \cos^{-1}\left(\frac{\pi}{2} - 2\pi\right) - 4\cos^2\left(\frac{\pi}{2} - 3\pi\right)$
5. $2 - \frac{\sin 8\alpha}{\sin^4 2\alpha - \cos^4 2\alpha}$
6. $\frac{2\cos^2 2\alpha - 1}{2\operatorname{tg}\left(\frac{\pi}{4} - 2\alpha\right)\sin^2\left(\frac{3\pi}{4} - 2\alpha\right)} - \operatorname{tg} 2\alpha + \cos 2\alpha - \sin 2\alpha$
7. $\frac{\operatorname{cotg} 2\alpha + \operatorname{tg} 2\alpha}{1 + \operatorname{tg} 4\alpha \operatorname{tg} 2\alpha}$
8. $1 - \sin^2\alpha - \sin^2\beta + 2\sin\alpha\sin\beta\cos(\alpha - \beta)$
9. $1 + \cos\left(2\pi - \frac{3\pi}{2}\right) + \sin\left(2\alpha + \frac{3\pi}{2}\right) - \operatorname{cotg}\left(\frac{\pi}{2} + 2\alpha\right)$
10. $4\cos^2\left(2\alpha - \frac{3\pi}{2}\right) + \cos(2\alpha - \pi) + \sin\left(\frac{5\pi}{2} - 6\alpha\right)$
11. $\frac{\sqrt{1 + \sin\alpha} + \sqrt{1 - \sin\alpha}}{\sqrt{1 + \sin\alpha} - \sqrt{1 - \sin\alpha}}$, ако а) $0^\circ \lt \alpha \lt 90^\circ$; б) $90^\circ \lt \alpha \lt 180^\circ$
12. $2\sin^2\alpha + \sqrt{3}\sin^4\alpha - \frac{4\operatorname{tg} 2\alpha\,(1 - \operatorname{tg}^2 2\alpha)}{\sin 8\alpha\,(1 + \operatorname{tg}^2 2\alpha)}$
13. $\cos^2(\alpha - 2\beta) - \cos^2\left(\alpha - \frac{\pi}{2}\right) - \cos^2(2\beta - \alpha)$
14. $1 - \cos(\pi - 8\alpha) - \cos(\pi + 4\alpha)$
15. $\cos 2\alpha - \sin 4\alpha - \cos 6\alpha$
16. $\sin^2\left(\frac{\pi}{2} + \alpha\right) + \cos^2\left(\frac{\pi}{2} + \alpha\right) - \cos\left(\alpha - \frac{3\pi}{2}\right) + \sin\left(\frac{3\pi}{2} + \alpha\right)$
17. $2\cos^2 2\alpha + \sqrt{3}\sin 4\alpha - 1$