Wiedząc, ze \(\tan \alpha = 2/3\) i \(\alpha\) jest kątem ostrym oblicz wartość wyrażenia:
a) \(\sin\alpha\)
b) \(\cos2\alpha\),
c) \(\frac{\sin\alpha - 4\cos\alpha}{2\sin \alpha + \cos \alpha} \)
d) \(\sin^2 \alpha + \cos^2 \alpha - \frac{1}{\cos^2 \alpha}\)
Zadanie z trygonometrii
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Jerry
- Expert
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- Rejestracja: 18 maja 2009, 09:23
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Re: Zadanie z trygonometrii
Skoro \(\alpha\) jest ostry i \(\tg\alpha={2\over3}\), to \(\begin{cases}\sin\alpha={2\over\sqrt{2^2+3^2}}\\\cos\alpha={3\over\sqrt{2^2+3^2}}\end{cases}\) i
a) \(\sin\alpha={2\over\sqrt{13}}=\ldots\)
b) \(\cos2\alpha=\cos^2\alpha-\sin^2\alpha=\left({3\over\sqrt{13}}\right)^2-\left({2\over\sqrt{13}}\right)^2=\ldots\)
c) \(\dfrac{\sin\alpha-4\cos\alpha}{2\sin\alpha+\cos\alpha}=\dfrac{\tg\alpha-4\cdot1}{2\tg\alpha+1}=\dfrac{{2\over3}-4\cdot1}{2\cdot{2\over3}+1}=\ldots\)
d) \(\sin^2\alpha+\cos^2\alpha-\frac{1}{\cos^2\alpha}=1-\left(\frac{\sqrt{13}}{3}\right)^2=\ldots\)
Pozdrawiam
a) \(\sin\alpha={2\over\sqrt{13}}=\ldots\)
b) \(\cos2\alpha=\cos^2\alpha-\sin^2\alpha=\left({3\over\sqrt{13}}\right)^2-\left({2\over\sqrt{13}}\right)^2=\ldots\)
c) \(\dfrac{\sin\alpha-4\cos\alpha}{2\sin\alpha+\cos\alpha}=\dfrac{\tg\alpha-4\cdot1}{2\tg\alpha+1}=\dfrac{{2\over3}-4\cdot1}{2\cdot{2\over3}+1}=\ldots\)
d) \(\sin^2\alpha+\cos^2\alpha-\frac{1}{\cos^2\alpha}=1-\left(\frac{\sqrt{13}}{3}\right)^2=\ldots\)
Pozdrawiam
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janusz55
- Fachowiec
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Re: Zadanie z trygonometrii
Albo
\( (a) \ \ \begin{cases} \tg(\alpha) = \frac{\sin(\alpha)}{\cos(\alpha)}= \frac{2}{3} \\ \sin^2(\alpha) + \cos^2(\alpha) = 1 \end{cases}.\)
\(\sin^2(\alpha) = 1 - \cos^2(\alpha) = 1 - \left(\frac{3}{2}\sin(\alpha)\right)^2 = 1 -\frac{9}{4}\sin^2(\alpha) \)
Stąd
\( \frac{4}{4}\sin^2(\alpha) + \frac{9}{4}\sin^2(\alpha) = 1.\)
\( \frac{13}{4}\sin^2(\alpha) = 1\)
\( \sin^2(\alpha) = \frac{4}{13}. \)
\( \sin(\alpha) = \frac{2}{\sqrt{13}} = \frac{2\sqrt{13}}{13}.\)
\( \cos^2(\alpha) = 1 - \sin^2(\alpha) = 1 - \frac{4}{13} = \frac{9}{13}.\)
\( \cos(\alpha) = \frac{3}{\sqrt{13}} = \frac{3\sqrt{13}}{13}.\)
\( (b) \ \ \cos(2\alpha) = \cos^2(\alpha) -\sin^2(\alpha) = \frac{9}{13} - \frac{4}{13} = \frac{5}{13}.\)
\( (c) \ \ \frac{\sin(\alpha) -4\cos(\alpha)}{2\sin(\alpha) +\cos(\alpha)} =\frac{\frac{2}{\sqrt{13}} - 4\cdot \frac{3}{\sqrt{13}}}{ 2\cdot \frac{2}{\sqrt{13}}+ \frac{3}{\sqrt{13}}} = -\frac{10}{7}.\)
\( (d) \ \ \sin^2(\alpha) + \cos^2(\alpha) - \frac{1}{\cos^2(\alpha)} = 1 - \frac{13}{9} = -\frac{4}{9}.\)
\( (a) \ \ \begin{cases} \tg(\alpha) = \frac{\sin(\alpha)}{\cos(\alpha)}= \frac{2}{3} \\ \sin^2(\alpha) + \cos^2(\alpha) = 1 \end{cases}.\)
\(\sin^2(\alpha) = 1 - \cos^2(\alpha) = 1 - \left(\frac{3}{2}\sin(\alpha)\right)^2 = 1 -\frac{9}{4}\sin^2(\alpha) \)
Stąd
\( \frac{4}{4}\sin^2(\alpha) + \frac{9}{4}\sin^2(\alpha) = 1.\)
\( \frac{13}{4}\sin^2(\alpha) = 1\)
\( \sin^2(\alpha) = \frac{4}{13}. \)
\( \sin(\alpha) = \frac{2}{\sqrt{13}} = \frac{2\sqrt{13}}{13}.\)
\( \cos^2(\alpha) = 1 - \sin^2(\alpha) = 1 - \frac{4}{13} = \frac{9}{13}.\)
\( \cos(\alpha) = \frac{3}{\sqrt{13}} = \frac{3\sqrt{13}}{13}.\)
\( (b) \ \ \cos(2\alpha) = \cos^2(\alpha) -\sin^2(\alpha) = \frac{9}{13} - \frac{4}{13} = \frac{5}{13}.\)
\( (c) \ \ \frac{\sin(\alpha) -4\cos(\alpha)}{2\sin(\alpha) +\cos(\alpha)} =\frac{\frac{2}{\sqrt{13}} - 4\cdot \frac{3}{\sqrt{13}}}{ 2\cdot \frac{2}{\sqrt{13}}+ \frac{3}{\sqrt{13}}} = -\frac{10}{7}.\)
\( (d) \ \ \sin^2(\alpha) + \cos^2(\alpha) - \frac{1}{\cos^2(\alpha)} = 1 - \frac{13}{9} = -\frac{4}{9}.\)