Równiania i nierówności trygonometryczne

[AZS06]

Zależy mi na sprawdzeniu, moich obliczeń i wyników.

1. Rozwiąż równania.
a)\(\displaystyle{ \sin\frac{1}{3}x = \frac{\sqrt2}{2}}\)
\(\displaystyle{ t =\frac{1}{3}x}\)
\(\displaystyle{ \sin t = \frac{\sqrt{2}}{2}}\)
\(\displaystyle{ \sin t = \frac{\sqrt{2}}{2}}\)
\(\displaystyle{ t_1 = \frac{\pi}{4} + 2k\pi}\) \(\displaystyle{ \vee}\) \(\displaystyle{ t_2 = \frac{3}{4}\pi + 2k\pi}\)
\(\displaystyle{ \frac{1}{3}x_1 = \frac{\pi}{4} + 2k\pi}\) \(\displaystyle{ \vee}\) \(\displaystyle{ \frac{1}{3}x_2 = \frac{3}{4}\pi + 2k\pi}\)
\(\displaystyle{ x_1 = \frac{3}{4}\pi + 6k\pi}\) \(\displaystyle{ \vee}\) \(\displaystyle{ x_2 = \frac{9}{4}\pi + 6k\pi}\)

b)\(\displaystyle{ \sin x + \cos^2 x +1 =0}\)
\(\displaystyle{ \sin x +1 -\sin^2 x +1 = 0}\)
\(\displaystyle{ -\sin^2 x + \sin x +2 = 0}\)
\(\displaystyle{ \sin x = t}\)
\(\displaystyle{ -t^2 + t +2 =0}\)
\(\displaystyle{ \Delta = 9}\)
\(\displaystyle{ t_1 = -2, \notin Y}\) \(\displaystyle{ \wedge}\) \(\displaystyle{ t_2 = -1, \in Y}\)
\(\displaystyle{ \sin x = -1}\)
\(\displaystyle{ \ x _1= -\frac{\pi}{2} + 2k\pi}\)

c) \(\displaystyle{ tg \frac{1}{4}x = \frac{\sqrt3}{3}}\)
\(\displaystyle{ \frac{1}{4}x = t}\)
\(\displaystyle{ \tg t = \frac{\sqrt3}{3}}\)
\(\displaystyle{ t_1 = \frac{\pi}{6} + k\pi}\)
\(\displaystyle{ \frac{1}{4}x = \frac{\pi}{6} + k\pi}\)
\(\displaystyle{ x = \frac{2}{3} \pi + 4k\pi}\)

d) \(\displaystyle{ \sin^2 x + \frac{1}{4} = \sin x}\)
\(\displaystyle{ \sin^2 x - \sin x + \frac{1}{4} = 0}\)
\(\displaystyle{ \sin x = t}\)
\(\displaystyle{ t^2 - t + \frac{1}{4} = 0}\)
\(\displaystyle{ \Delta = 0}\)
\(\displaystyle{ t_{1,2} = \frac{1}{2}}\)
\(\displaystyle{ \sin x = \frac {1}{2}}\)
\(\displaystyle{ x_1 = \frac{\pi}{6} + 2k\pi}\) \(\displaystyle{ \vee}\) \(\displaystyle{ x_2 = \frac{5}{6}\pi + 2k\pi}\)

2. Rozwiąż nierówność
a)\(\displaystyle{ \tg x \leq -\sqrt3, x \in (-\frac{\pi}{2}; \frac{\pi}{2})}\)
\(\displaystyle{ ....}\)
\(\displaystyle{ x \in ( -\frac{\pi}{2}; - \frac{\pi}{3}>}\)

b)\(\displaystyle{ \ctg x \leq -\sqrt3, x \in (0; \pi)}\)
\(\displaystyle{ ....}\)
\(\displaystyle{ x \in <\frac{5}{6}\pi; \pi )}\)

3. Oblicz bez użycia tablic matematycznych.
a)\(\displaystyle{ 3\cos(-300^ \circ )*\sin405^\circ tg135^\circ =}\)
\(\displaystyle{ = 3\cos(270^ \circ + 30^ \circ )*\sin(360^\circ + 45^\circ) tg(90^\circ + 45^\circ) =}\)
\(\displaystyle{ = 3\sin30^ \circ*\sin45^\circ* (-ctg)45^\circ =}\)
\(\displaystyle{ = 3 * \frac{1}{2}* \frac{\sqrt2}{2}* -1= -\frac{3\sqrt2}{4}}\)

b)\(\displaystyle{ 2\sin(-225^ \circ )*\ctg330^\circ tg405^\circ =}\)
\(\displaystyle{ = 2 (-\sin)(180^ \circ + 45^\circ )*\ctg(360^\circ - 30^\circ) tg(360^\circ + 45^\circ) =}\)
\(\displaystyle{ = 2 \sin45^\circ *(-\ctg)30^\circ* tg45^\circ =}\)
\(\displaystyle{ = \sqrt2 *(-\sqrt3)* 1= -\sqrt6}\)

4. Wyznacz wartości pozostałych funkcji trygonometrycznych wiedząc, że:
a) \(\displaystyle{ \tg x = -\frac{3}{4}, x \in (\frac{\pi}{2};\pi)}\)
\(\displaystyle{ \ctg x = -\frac{4}{3}}\)
\(\displaystyle{ \left\{ \begin{array}{l} \tg x = \frac{\sin x}{\cos x}\\sin^2 x + cos^2 x = 1\end{array}}\)
\(\displaystyle{ \left\{ \begin{array}{l} -3\cos x = 4\sin x\\sin^2 x + cos^2 x = 1\end{array}}\)
\(\displaystyle{ \left\{ \begin{array}{l} \cos x =-\frac{4}{3}\sin x\\sin^2 x +\frac{16}{9} \sin^2 x = 1\end{array}}\)
\(\displaystyle{ \left\{ \begin{array}{l} \cos x =-\frac{4}{3}\sin x\\sin^2 x = \frac{9}{25}\end{array}}\)
\(\displaystyle{ \left\{ \begin{array}{l} \cos x =-\frac{4}{3}\sin x\\sin x = \frac{3}{5}, \in Df \vee \sin x = -\frac{3}{5}, \notin Df \end{array}}\)
\(\displaystyle{ \left\{ \begin{array}{l} \sin x =\frac{3}{5} \\cos x = -\frac{4}{5} \end{array}}\)

b) \(\displaystyle{ c\tg x = -0,2 , x \in (\frac{\pi}{2};\pi)}\)
\(\displaystyle{ \tg x = -5}\)
\(\displaystyle{ \left\{ \begin{array}{l} \tg x = \frac{\sin x}{\cos x}\\sin^2 x + cos^2 x = 1\end{array}}\)
\(\displaystyle{ \left\{ \begin{array}{l} \sin x = -5\cos x\\(-5\cos^2 x) + cos^2 x = 1\end{array}}\)
\(\displaystyle{ \left\{ \begin{array}{l} \sin x = -5\cos x\\ \cos^2 x = \frac{1}{26}\end{array}}\)
\(\displaystyle{ \left\{ \begin{array}{l} \sin x = -5\cos x\\ \cos x = \frac{\sqrt{26}}{26}, \notin Df \vee \cos x = -\frac{\sqrt{26}}{26}, \in Df \end{array}}\)
\(\displaystyle{ \left\{ \begin{array}{l} \cos x = -\frac{\sqrt{26}}{26}\\ \sin x = \frac{(-5)*(-\sqrt{26})}{26}\end{array}}\)
\(\displaystyle{ \left\{ \begin{array}{l} \cos x = -\frac{\sqrt{26}}{26}\\ \sin x = \frac{5\sqrt{26}}{26}\end{array}}\)

5. Sprawdź
\(\displaystyle{ (\frac{1}{\sin x} + \frac{1}{\cos x})(\cos x - \sin x) = \ctg x - \tg x}\)

\(\displaystyle{ \sin x \neq 0, \cos x \neq 0}\)

\(\displaystyle{ L = \frac{(\cos x + \sin x)(\cos x - sin x)}{\sin x \cos x} =}\)
\(\displaystyle{ = \frac{\cos^2 x - \sin^2 x}{\sin x \cos x} =}\)
\(\displaystyle{ = \frac{\cos^2 x}{\sin x \cos x} - \frac{\sin^2 x}{\sin x \cos x} = \ctg x - \tg x = P}\)

[mmoonniiaa]

1. b) \(\displaystyle{ t_1=2 \notin Y}\) Ale wyniku to oczywiście nie zmienia -- 29 stycznia 2009, 13:37 --Oprócz tego malutkiego błędu, wszystko rozwiązane poprawnie.

[AZS06]

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