Oblicz wartości pozostałych funkcjie; podaj miarę łukową.

[maniek69]

Oblicz wartości pozostałych funkcji trygonometrycznych kąta\(\displaystyle{ \alpha}\) wiedząc, że:
a) \(\displaystyle{ \sin\alpha=0,6}\) i \(\displaystyle{ \alpha}\) jest kątem ostrym
b) \(\displaystyle{ \cos\alpha=\frac{1}{7}}\) i \(\displaystyle{ \alpha \in \left( \frac{3}{2} \pi ; 2\pi\right)}\)
c) \(\displaystyle{ \tan\alpha=2}\) i \(\displaystyle{ \alpha \in \left(\pi ; \frac{3}{2} \pi \right)}\)



2)Podaj miarę łukową kąta ostrego \(\displaystyle{ \alpha}\) , jeżeli
a) \(\displaystyle{ \sin \alpha=\frac{ \sqrt{3} }{2}}\)
b) \(\displaystyle{ \cos\alpha=\frac{ \sqrt{5} }{ \sqrt{10} }}\)
c) \(\displaystyle{ \tan\alpha=\frac{1+ \sqrt{3} }{3+ \sqrt{3} }}\)



Baaardzo proszę o rozwiązanie tych zadań, mam z nich sprawdzian, a nie kumam czaczy:((((

[AZS06]

Zadanie 1.
a) \(\displaystyle{ \sin\alpha = \frac{6}{10}}\)
\(\displaystyle{ \alpha \in I cw}\) więc \(\displaystyle{ \sin\alpha, \cos\alpha, \tg\alpha, \ctg\alpha > 0}\)
\(\displaystyle{ \cos \alpha = \frac{1}{\sin \alpha}}\)

\(\displaystyle{ \cos \alpha = \frac{10}{6}}\)

\(\displaystyle{ tg \alpha = \frac{\sin\alpha}{\cos\alpha}}\)

\(\displaystyle{ ctg \alpha = \frac{\cos\alpha}{\sin\alpha}}\)

b) analogicznie do a, z tym, że:
\(\displaystyle{ \alpha \in IV cw}\)
więc
\(\displaystyle{ \sin\alpha, \tg\alpha, \ctg\alpha <0}\)
\(\displaystyle{ \cos\alpha >0}\)

c)

\(\displaystyle{ \alpha \in III cw}\)
więc
\(\displaystyle{ \sin\alpha, \cos\alpha <0}\)
\(\displaystyle{ \tg\alpha, \ctg\alpha >0}\)

\(\displaystyle{ \left\{ \begin{array}{l} \tg \alpha = \frac{\sin\alpha}{\cos\alpha}\\sin^2 \alpha + cos^2 \alpha = 1\end{array}}\)

\(\displaystyle{ \left\{ \begin{array}{l} 2 = \frac{\sin\alpha}{\cos\alpha}\\sin^2 \alpha + cos^2 \alpha = 1\end{array}}\)

\(\displaystyle{ \left\{ \begin{array}{l} 2\cos\alpha = \sin\alpha\\5cos^2 \alpha = 1\end{array}}\)

\(\displaystyle{ \left\{ \begin{array}{l} 2\cos\alpha = \sin\alpha\\cos^2 \alpha = \frac{1}{5}\end{array}}\)

\(\displaystyle{ \left\{ \begin{array}{l} 2\cos\alpha = \sin\alpha\\cos^2 \alpha = \frac{1}{5}\end{array}}\)

\(\displaystyle{ \left\{ \begin{array}{l} 2\cos\alpha = \sin\alpha\\cos \alpha = \frac{1}{\sqrt5}\end{array}}\)

\(\displaystyle{ \left\{ \begin{array}{l} 2\cos\alpha = \sin\alpha\\cos \alpha = \frac{\sqrt5}{5} \notin Df \wedge \cos \alpha = -\frac{\sqrt5}{5} \in Df\end{array}}\)

\(\displaystyle{ \left\{ \begin{array}{l} \sin\alpha = 2\cos\alpha \\cos \alpha = -\frac{\sqrt5}{5} \end{array}}\)

\(\displaystyle{ \left\{ \begin{array}{l} \sin\alpha = -\frac{2\sqrt5}{5} \\cos \alpha = -\frac{\sqrt5}{5} \end{array}}\)

\(\displaystyle{ \ctg \alpha = \frac{1}{\tg \alpha}}\)

[AZS06]

Zadanie 2
a)
\(\displaystyle{ \sin\alpha = \frac{\sqrt3}{2}}\)

\(\displaystyle{ \sin60^\circ = \frac{\sqrt3}{2}}\)

\(\displaystyle{ \alpha = 60^\circ}\)

\(\displaystyle{ \alpha = \frac{\pi}{3}}\)

b)

\(\displaystyle{ \cos\alpha = \frac{\sqrt5}{\sqrt10}}\)
\(\displaystyle{ \cos\alpha = \frac{\sqrt50}{10}}\)


\(\displaystyle{ \cos\alpha = \frac{5\sqrt2}{10}}\)
\(\displaystyle{ \cos\alpha = \frac{\sqrt2}{2}}\)

\(\displaystyle{ \cos45^\circ = \frac{\sqrt2}{2}}\)

\(\displaystyle{ \alpha = 45^\circ}\)

\(\displaystyle{ \alpha = \frac{\pi}{4}}\)

c)

\(\displaystyle{ \tg\alpha = \frac{1+\sqrt3}{3+\sqrt3}}\)

\(\displaystyle{ \tg\alpha = \frac{1+\sqrt3}{3+\sqrt3}*\frac{3-\sqrt3}{3-\sqrt3}}\)

\(\displaystyle{ \tg\alpha = \frac{3 - \sqrt3 + 3\sqrt3 - 3}{9 - 3}}\)

\(\displaystyle{ \tg\alpha = \frac{2\sqrt3}{6}}\)

\(\displaystyle{ \tg\alpha = \frac{\sqrt3}{3}}\)

\(\displaystyle{ \tg30^\circ = \frac{\sqrt3}{3}}\)

\(\displaystyle{ \alpha = 30^\circ}\)

\(\displaystyle{ \alpha = \frac{\pi}{6}}\)

POZDRAWIAM !!
Mam nadzieje, ze pomogłem