Trygonometria, zadanie nr 548
ostatnie wiadomo艣ci | regulamin | latex
| Autor | Zadanie / Rozwi膮zanie |
pawelek post贸w: 10 |  2011-01-30 10:46:21Witam serdecznie, czy jest kto艣 w stanie rozwi膮za膰 mi poni偶sze zadanie? Sprawd藕, czy podane r贸wno艣ci s膮 to偶samo艣ciami trygonometrycznymi. Podaj konieczne za艂o偶enia. a) $\cos \alpha \cdot \frac{\tg \alpha}{\sin \alpha} = 1$ b) $1-2 \sin^2 \alpha = 2\cos^2 \alpha - 1$ c) $\frac{\sin \alpha + \tg \alpha}{\sin \alpha} = 1 + \frac{1}{\cos \alpha}$ d) $\frac{\sin \alpha + \cos \alpha}{\cos \alpha} = 1 + \tg \alpha$ e) $\left( \cos^2 \alpha - 1 \right) \left( \tg^2 \alpha + 1 \right) + \tg^2 \alpha = 0$ f) $\left( \frac{1}{\sin \alpha} - \frac {1}{\cos \alpha} \right) \left( \sin \alpha + \cos \alpha \right) = \ctg \alpha - \tg \alpha$ g) $\left( 1 - \cos \alpha \right) \left( 1 + \sin \alpha \right) = \cos^2 \alpha - 1$ h) $\frac{ 1 - \cos \alpha}{ 1 + \cos \alpha} - \frac{ 1 + \cos \alpha}{ 1 - \cos \alpha} = - \frac{2}{\sin \alpha}$ i) $\frac{2}{\cos^2 \alpha} - 1 = 1 + 2 \tg^2 \alpha$ j) $\frac{1}{1 - \sin \alpha} + \frac{1}{1 + \sin \alpha} = \frac{2}{\cos^2 \alpha}$ Z g贸ry dzi臋kuje za rozwi膮zanie. Pozdrawiam, pawelek |
jarah post贸w: 448 | 2011-01-30 11:32:08 $\cos \alpha \cdot \frac{tg \alpha}{\sin \alpha} =\cos \alpha \cdot \frac{sin\alpha}{cos\alpha}\cdot\frac{1}{\sin \alpha} = 1$ |
jarah post贸w: 448 | 2011-01-30 11:37:15 $1-2 \sin^2 \alpha =sin^{2}\alpha+cos^{2}\alpha-2sin^{2}\alpha=cos^{2}\alpha-sin^{2}\alpha=cos^{2}\alpha-(1-cos^{2}\alpha)=2\cos^2 \alpha - 1$ |
jarah post贸w: 448 | 2011-01-30 11:39:50 $\frac{\sin \alpha + tg \alpha}{\sin \alpha} = 1+\frac{tg\alpha}{sin\alpha}= 1+\frac{sin\alpha}{cos\alpha}\cdot\frac{1}{sin\alpha}= 1 + \frac{1}{\cos \alpha}$ |
jarah post贸w: 448 | 2011-01-30 11:41:48 $\frac{\sin \alpha + \cos \alpha}{\cos \alpha}=\frac{cos\alpha}{cos\alpha}+\frac{sin\alpha}{cos\alpha} = 1 + tg \alpha$ |
jarah post贸w: 448 | 2011-01-30 11:44:19 $\left( \cos^2 \alpha - 1 \right) \left( tg^2 \alpha + 1 \right) + tg^2 \alpha =sin^{2}\alpha+cos^{2}\alpha-tg^{2}\alpha-1+tg^{2}\alpha=1-1= 0$ |
jarah post贸w: 448 | 2011-01-30 11:46:07 $\left( \frac{1}{\sin \alpha} - \frac {1}{\cos \alpha} \right) \left( \sin \alpha + \cos \alpha \right)=1+ctg\alpha-tg\alpha-1 = ctg \alpha - tg \alpha$ |
jarah post贸w: 448 | 2011-01-30 12:34:25 $\frac{1}{1 - \sin \alpha} + \frac{1}{1 + \sin \alpha} = \frac{1 + \sin \alpha}{(1 - \sin \alpha)(1 + \sin \alpha)} + \frac{1 - \sin \alpha}{(1 + \sin \alpha)(1 - \sin \alpha)}=\frac{2}{1-sin^{2}\alpha}=\frac{2}{\cos^2 \alpha}$ |
jarah post贸w: 448 | 2011-01-30 12:38:27 $\frac{2}{\cos^2 \alpha} - 1 = \frac{2sin^{2}\alpha+2cos^{2}\alpha}{cos^{2}\alpha}-1=2+2 tg^2 \alpha-1=1 + 2 tg^2 \alpha$ |
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