1. El problema pide calcular las funciones trigonométricas del ángulo α para los incisos 3, 6, 9 y 12. 2. Recordemos las funciones trigonométricas básicas: - $$\sin \alpha = \frac{y}{r}$$ - $$\cos \alpha = \frac{x}{r}$$ - $$\tan \alpha = \frac{y}{x}$$ - $$\csc \alpha = \frac{1}{\sin \alpha}$$ - $$\sec \alpha = \frac{1}{\cos \alpha}$$ - $$\cot \alpha = \frac{1}{\tan \alpha}$$ Donde $r = \sqrt{x^2 + y^2}$ es la distancia del punto $P(x,y)$ al origen. 3. Inciso 3: $\sin \alpha = -\frac{1}{2}$ y $\alpha$ está en el 3er cuadrante. - En el 3er cuadrante, $\sin$ y $\cos$ son negativos, $\tan$ es positivo. - Calculamos $\cos \alpha$ usando la identidad $\sin^2 \alpha + \cos^2 \alpha = 1$: $$\cos \alpha = -\sqrt{1 - \sin^2 \alpha} = -\sqrt{1 - \left(-\frac{1}{2}\right)^2} = -\sqrt{1 - \frac{1}{4}} = -\sqrt{\frac{3}{4}} = -\frac{\sqrt{3}}{2}$$ - Calculamos $\tan \alpha = \frac{\sin \alpha}{\cos \alpha} = \frac{-\frac{1}{2}}{-\frac{\sqrt{3}}{2}} = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3}$ - Calculamos $\csc \alpha = \frac{1}{\sin \alpha} = -2$ - Calculamos $\sec \alpha = \frac{1}{\cos \alpha} = -\frac{2}{\sqrt{3}} = -\frac{2\sqrt{3}}{3}$ - Calculamos $\cot \alpha = \frac{1}{\tan \alpha} = \sqrt{3}$ 4. Inciso 6: $P(7,24)$ - Calculamos $r = \sqrt{7^2 + 24^2} = \sqrt{49 + 576} = \sqrt{625} = 25$ - Calculamos las funciones: $$\sin \alpha = \frac{24}{25}$$ $$\cos \alpha = \frac{7}{25}$$ $$\tan \alpha = \frac{24}{7}$$ $$\csc \alpha = \frac{25}{24}$$ $$\sec \alpha = \frac{25}{7}$$ $$\cot \alpha = \frac{7}{24}$$ 5. Inciso 9: $P(3,2)$ - Calculamos $r = \sqrt{3^2 + 2^2} = \sqrt{9 + 4} = \sqrt{13}$ - Calculamos las funciones: $$\sin \alpha = \frac{2}{\sqrt{13}} = \frac{2\sqrt{13}}{13}$$ $$\cos \alpha = \frac{3}{\sqrt{13}} = \frac{3\sqrt{13}}{13}$$ $$\tan \alpha = \frac{2}{3}$$ $$\csc \alpha = \frac{\sqrt{13}}{2}$$ $$\sec \alpha = \frac{\sqrt{13}}{3}$$ $$\cot \alpha = \frac{3}{2}$$ 6. Inciso 12: $P(2,2\sqrt{2})$ - Calculamos $r = \sqrt{2^2 + (2\sqrt{2})^2} = \sqrt{4 + 8} = \sqrt{12} = 2\sqrt{3}$ - Calculamos las funciones: $$\sin \alpha = \frac{2\sqrt{2}}{2\sqrt{3}} = \frac{\sqrt{2}}{\sqrt{3}} = \frac{\sqrt{6}}{3}$$ $$\cos \alpha = \frac{2}{2\sqrt{3}} = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3}$$ $$\tan \alpha = \frac{2\sqrt{2}}{2} = \sqrt{2}$$ $$\csc \alpha = \frac{1}{\sin \alpha} = \frac{3}{\sqrt{6}} = \frac{3\sqrt{6}}{6} = \frac{\sqrt{6}}{2}$$ $$\sec \alpha = \frac{1}{\cos \alpha} = \sqrt{3}$$ $$\cot \alpha = \frac{1}{\tan \alpha} = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}$$ **Respuesta final:** - Inciso 3: $\sin \alpha = -\frac{1}{2}$, $\cos \alpha = -\frac{\sqrt{3}}{2}$, $\tan \alpha = \frac{\sqrt{3}}{3}$, $\csc \alpha = -2$, $\sec \alpha = -\frac{2\sqrt{3}}{3}$, $\cot \alpha = \sqrt{3}$ - Inciso 6: $\sin \alpha = \frac{24}{25}$, $\cos \alpha = \frac{7}{25}$, $\tan \alpha = \frac{24}{7}$, $\csc \alpha = \frac{25}{24}$, $\sec \alpha = \frac{25}{7}$, $\cot \alpha = \frac{7}{24}$ - Inciso 9: $\sin \alpha = \frac{2\sqrt{13}}{13}$, $\cos \alpha = \frac{3\sqrt{13}}{13}$, $\tan \alpha = \frac{2}{3}$, $\csc \alpha = \frac{\sqrt{13}}{2}$, $\sec \alpha = \frac{\sqrt{13}}{3}$, $\cot \alpha = \frac{3}{2}$ - Inciso 12: $\sin \alpha = \frac{\sqrt{6}}{3}$, $\cos \alpha = \frac{\sqrt{3}}{3}$, $\tan \alpha = \sqrt{2}$, $\csc \alpha = \frac{\sqrt{6}}{2}$, $\sec \alpha = \sqrt{3}$, $\cot \alpha = \frac{\sqrt{2}}{2}$