Задачи по тригонометрия

Ъглите $\alpha$ и $\beta$ са от I квадрант, затова $\cos\alpha \gt 0$ и $\sin\beta \gt 0$:

$\cos\alpha = \sqrt{1 - \sin^2\alpha} = \sqrt{1 - \frac{5}{25}} = \sqrt{\frac{20}{25}} = \frac{2\sqrt{5}}{5}$

$\sin\beta = \sqrt{1 - \cos^2\beta} = \sqrt{1 - \frac{10}{100}} = \sqrt{\frac{90}{100}} = \frac{3\sqrt{10}}{10}$

$\sin(\beta - \alpha) = \sin\beta\cos\alpha - \cos\beta\sin\alpha = \frac{3\sqrt{10}}{10} \cdot \frac{2\sqrt{5}}{5} - \frac{\sqrt{10}}{10} \cdot \frac{\sqrt{5}}{5}$

$= \frac{6\sqrt{50}}{50} - \frac{\sqrt{50}}{50} = \frac{5\sqrt{50}}{50} = \frac{25\sqrt{2}}{50} = \frac{\sqrt{2}}{2}$

От $0 \lt \alpha \lt \frac{\pi}{2}$ и $0 \lt \beta \lt \frac{\pi}{2}$ следва $-\frac{\pi}{2} \lt \beta - \alpha \lt \frac{\pi}{2}$. В този интервал единственият ъгъл със синус $\frac{\sqrt{2}}{2}$ е $\frac{\pi}{4}$, следователно $\beta - \alpha = \frac{\pi}{4}$.

$\cos\alpha = \sqrt{1 - \sin^2\alpha} = \sqrt{1 - \frac{225}{289}} = \sqrt{\frac{64}{289}} = \frac{8}{17}$

$\cos\beta = \sqrt{1 - \sin^2\beta} = \sqrt{1 - \frac{64}{289}} = \sqrt{\frac{225}{289}} = \frac{15}{17}$

$\sin(\alpha + \beta) = \sin\alpha\cos\beta + \cos\alpha\sin\beta = \frac{15}{17} \cdot \frac{15}{17} + \frac{8}{17} \cdot \frac{8}{17}$

$= \frac{225}{289} + \frac{64}{289} = \frac{289}{289} = 1$

От $0 \lt \alpha + \beta \lt \pi$ и $\sin(\alpha + \beta) = 1$ следва $\alpha + \beta = \frac{\pi}{2}$.

$\cos\alpha = \sqrt{1 - \frac{1}{9}} = \sqrt{\frac{8}{9}} = \frac{2\sqrt{2}}{3}$

$\cos\beta = \sqrt{1 - \frac{11}{33^2}} = \sqrt{\frac{1078}{1089}} = \sqrt{\frac{49 \cdot 22}{1089}} = \frac{7\sqrt{22}}{33}$

$\cos\gamma = \sqrt{1 - \frac{9 \cdot 11}{11^2}} = \sqrt{\frac{22}{121}} = \frac{\sqrt{22}}{11}$

Пресмятаме синуса и косинуса на $\alpha + \beta$:

$\sin(\alpha + \beta) = \sin\alpha\cos\beta + \cos\alpha\sin\beta = \frac{1}{3} \cdot \frac{7\sqrt{22}}{33} + \frac{2\sqrt{2}}{3} \cdot \frac{\sqrt{11}}{33}$

$= \frac{7\sqrt{22}}{99} + \frac{2\sqrt{22}}{99} = \frac{9\sqrt{22}}{99} = \frac{\sqrt{22}}{11}$

$\cos(\alpha + \beta) = \cos\alpha\cos\beta - \sin\alpha\sin\beta = \frac{2\sqrt{2}}{3} \cdot \frac{7\sqrt{22}}{33} - \frac{1}{3} \cdot \frac{\sqrt{11}}{33}$

$= \frac{28\sqrt{11}}{99} - \frac{\sqrt{11}}{99} = \frac{27\sqrt{11}}{99} = \frac{3\sqrt{11}}{11}$

Тогава

$\sin(\alpha + \beta + \gamma) = \sin(\alpha + \beta)\cos\gamma + \cos(\alpha + \beta)\sin\gamma = \frac{\sqrt{22}}{11} \cdot \frac{\sqrt{22}}{11} + \frac{3\sqrt{11}}{11} \cdot \frac{3\sqrt{11}}{11}$

$= \frac{22}{121} + \frac{99}{121} = 1$

От $0 \lt \alpha + \beta + \gamma \lt \frac{3\pi}{2}$ и $\sin(\alpha + \beta + \gamma) = 1$ следва $\alpha + \beta + \gamma = \frac{\pi}{2}$.

$\alpha + \beta = \frac{\pi}{2} - \gamma \;\Rightarrow\; \operatorname{tg}(\alpha + \beta) = \operatorname{tg}\left(\frac{\pi}{2} - \gamma\right) = \operatorname{cotg}\gamma = \frac{1}{\operatorname{tg}\gamma}$

$\frac{\operatorname{tg}\alpha + \operatorname{tg}\beta}{1 - \operatorname{tg}\alpha \operatorname{tg}\beta} = \frac{1}{\operatorname{tg}\gamma} \;\Leftrightarrow\; \operatorname{tg}\alpha \operatorname{tg}\gamma + \operatorname{tg}\beta \operatorname{tg}\gamma = 1 - \operatorname{tg}\alpha \operatorname{tg}\beta$

$\Leftrightarrow\; \operatorname{tg}\alpha \operatorname{tg}\beta + \operatorname{tg}\beta \operatorname{tg}\gamma + \operatorname{tg}\alpha \operatorname{tg}\gamma = 1$

$\operatorname{cotg}(\alpha + \beta) = \operatorname{cotg}\left(\frac{\pi}{2} - \gamma\right) = \operatorname{tg}\gamma = \frac{1}{\operatorname{cotg}\gamma}$

$\frac{\operatorname{cotg}\alpha \operatorname{cotg}\beta - 1}{\operatorname{cotg}\alpha + \operatorname{cotg}\beta} = \frac{1}{\operatorname{cotg}\gamma} \;\Leftrightarrow\; \operatorname{cotg}\gamma\,(\operatorname{cotg}\alpha \operatorname{cotg}\beta - 1) = \operatorname{cotg}\alpha + \operatorname{cotg}\beta$

$\Leftrightarrow\; \operatorname{cotg}\alpha + \operatorname{cotg}\beta = \operatorname{cotg}\alpha \operatorname{cotg}\beta \operatorname{cotg}\gamma - \operatorname{cotg}\gamma$

$\Leftrightarrow\; \operatorname{cotg}\alpha + \operatorname{cotg}\beta + \operatorname{cotg}\gamma = \operatorname{cotg}\alpha \operatorname{cotg}\beta \operatorname{cotg}\gamma$

$\operatorname{tg}(\alpha + \beta) = \operatorname{tg}\frac{\pi}{4} = 1$

$\frac{\operatorname{tg}\alpha + \operatorname{tg}\beta}{1 - \operatorname{tg}\alpha \operatorname{tg}\beta} = 1 \;\Leftrightarrow\; \operatorname{tg}\alpha + \operatorname{tg}\beta = 1 - \operatorname{tg}\alpha \operatorname{tg}\beta \;\Leftrightarrow\; 1 + \operatorname{tg}\alpha + \operatorname{tg}\beta + \operatorname{tg}\alpha \operatorname{tg}\beta = 2$

$\Leftrightarrow\; (1 + \operatorname{tg}\alpha) + \operatorname{tg}\beta\,(1 + \operatorname{tg}\alpha) = 2 \;\Leftrightarrow\; (1 + \operatorname{tg}\alpha)(1 + \operatorname{tg}\beta) = 2$

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

Тъй като $\cos(\alpha - 60^\circ) \le 1$, то $\cos\alpha + \sqrt{3}\sin\alpha = 2\cos(\alpha - 60^\circ) \le 2$.

$\sqrt{2}(\sin\alpha + \cos\alpha) \le 2 \;\Leftrightarrow\; \frac{\sqrt{2}}{2}\sin\alpha + \frac{\sqrt{2}}{2}\cos\alpha \le 1$

$\Leftrightarrow\; \cos 45^\circ \sin\alpha + \sin 45^\circ \cos\alpha \le 1 \;\Leftrightarrow\; \sin(\alpha + 45^\circ) \le 1$

Последното неравенство е вярно за всяко $\alpha$. Преобразуванията са еквивалентни, затова е вярно и даденото неравенство.

$\sin 2\alpha \cos\beta + \cos 2\alpha \sin\beta = \sin(2\alpha + \beta) \le 1$

Използваме, че $|\cos\alpha| \le 1$ и $|\cos\beta| \le 1$:

$|\sin(\alpha + \beta)| = |\sin\alpha\cos\beta + \sin\beta\cos\alpha| \le |\sin\alpha|\,|\cos\beta| + |\sin\beta|\,|\cos\alpha| \le |\sin\alpha| + |\sin\beta|$

$\cos\alpha - \sin\alpha\sin\beta - \cos(\alpha + \beta) = \cos\alpha - \sin\alpha\sin\beta - \cos\alpha\cos\beta + \sin\alpha\sin\beta = \cos\alpha\,(1 - \cos\beta)$

За $\alpha \in$ I квадрант $\cos\alpha \gt 0$, а $1 - \cos\beta \ge 0$, защото $\cos\beta \le 1$. Следователно разликата е неотрицателна и $\cos(\alpha + \beta) \le \cos\alpha - \sin\alpha \sin\beta$. Равенство има само при $\cos\beta = 1$.

$\sqrt{3}\cos\alpha - \sin\alpha \le 2 \;\Leftrightarrow\; \frac{\sqrt{3}}{2}\cos\alpha - \frac{1}{2}\sin\alpha \le 1$

$\Leftrightarrow\; \sin 60^\circ \cos\alpha - \cos 60^\circ \sin\alpha \le 1 \;\Leftrightarrow\; \sin(60^\circ - \alpha) \le 1$

Последното неравенство е вярно за всяко $\alpha$, следователно е вярно и даденото.

$\sin\alpha - \cos\alpha \le \sqrt{2} \;\Leftrightarrow\; \frac{\sqrt{2}}{2}\sin\alpha - \frac{\sqrt{2}}{2}\cos\alpha \le 1$

$\Leftrightarrow\; \cos 45^\circ \sin\alpha - \sin 45^\circ \cos\alpha \le 1 \;\Leftrightarrow\; \sin(\alpha - 45^\circ) \le 1$

Последното неравенство е вярно за всяко $\alpha$, следователно е вярно и даденото.

$y = 2\left(\frac{\sqrt{3}}{2}\sin 2x - \frac{1}{2}\cos 2x\right)$

$= 2\left(\sin 2x \cos\frac{\pi}{6} - \cos 2x \sin\frac{\pi}{6}\right)$

$= 2\sin\left(2x - \frac{\pi}{6}\right)$

Тъй като $\sin\left(2x - \frac{\pi}{6}\right) \le 1$, то $y \le 2$. Стойността $2$ се достига, когато

$\sin\left(2x - \frac{\pi}{6}\right) = 1 \;\Leftrightarrow\; 2x - \frac{\pi}{6} = \frac{\pi}{2} + 2k\pi \;\Leftrightarrow\; x = \frac{\pi}{3} + k\pi, \; k \in \mathbb{Z}$

От тези стойности в интервала $0 \le x \le \pi$ е само $x = \frac{\pi}{3}$.

Отговор: НГС на $y$ е $2$ и се достига при $x = \frac{\pi}{3}$.