1. Problema: Determinar $\sin(\alpha + \beta)$, $\cos(\alpha + \beta)$, $\tan(\alpha - \beta)$ con $\sin \alpha = \frac{3}{5}$, $\alpha$ en $Q_1$, $\cos \beta = \frac{2\sqrt{5}}{5}$, $\beta$ en $Q_4$. Fórmulas: $$\sin(\alpha + \beta) = \sin \alpha \cos \beta + \cos \alpha \sin \beta$$ $$\cos(\alpha + \beta) = \cos \alpha \cos \beta - \sin \alpha \sin \beta$$ $$\tan(\alpha - \beta) = \frac{\tan \alpha - \tan \beta}{1 + \tan \alpha \tan \beta}$$ Paso 1: Hallar $\cos \alpha$ usando $\sin^2 \alpha + \cos^2 \alpha = 1$: $$\cos \alpha = \sqrt{1 - \sin^2 \alpha} = \sqrt{1 - \left(\frac{3}{5}\right)^2} = \sqrt{1 - \frac{9}{25}} = \sqrt{\frac{16}{25}} = \frac{4}{5}$$ Paso 2: Hallar $\sin \beta$ usando $\sin^2 \beta + \cos^2 \beta = 1$ y $\beta$ en $Q_4$ (donde $\sin \beta < 0$): $$\sin \beta = -\sqrt{1 - \cos^2 \beta} = -\sqrt{1 - \left(\frac{2\sqrt{5}}{5}\right)^2} = -\sqrt{1 - \frac{4 \cdot 5}{25}} = -\sqrt{1 - \frac{20}{25}} = -\sqrt{\frac{5}{25}} = -\frac{\sqrt{5}}{5}$$ Paso 3: Calcular $\sin(\alpha + \beta)$: $$\sin(\alpha + \beta) = \frac{3}{5} \cdot \frac{2\sqrt{5}}{5} + \frac{4}{5} \cdot \left(-\frac{\sqrt{5}}{5}\right) = \frac{6\sqrt{5}}{25} - \frac{4\sqrt{5}}{25} = \frac{2\sqrt{5}}{25}$$ Paso 4: Calcular $\cos(\alpha + \beta)$: $$\cos(\alpha + \beta) = \frac{4}{5} \cdot \frac{2\sqrt{5}}{5} - \frac{3}{5} \cdot \left(-\frac{\sqrt{5}}{5}\right) = \frac{8\sqrt{5}}{25} + \frac{3\sqrt{5}}{25} = \frac{11\sqrt{5}}{25}$$ Paso 5: Calcular $\tan \alpha = \frac{\sin \alpha}{\cos \alpha} = \frac{3/5}{4/5} = \frac{3}{4}$ Paso 6: Calcular $\tan \beta = \frac{\sin \beta}{\cos \beta} = \frac{-\frac{\sqrt{5}}{5}}{\frac{2\sqrt{5}}{5}} = -\frac{1}{2}$ Paso 7: Calcular $\tan(\alpha - \beta)$: $$\tan(\alpha - \beta) = \frac{\frac{3}{4} - (-\frac{1}{2})}{1 + \frac{3}{4} \cdot (-\frac{1}{2})} = \frac{\frac{3}{4} + \frac{1}{2}}{1 - \frac{3}{8}} = \frac{\frac{3}{4} + \frac{4}{8}}{\frac{5}{8}} = \frac{\frac{3}{4} + \frac{1}{2}}{\frac{5}{8}} = \frac{\frac{6}{8} + \frac{4}{8}}{\frac{5}{8}} = \frac{\frac{10}{8}}{\frac{5}{8}} = \frac{10}{8} \cdot \frac{8}{5} = 2$$ --- 2. Problema: Determinar $\sin(\alpha + \beta)$, $\cos(\alpha + \beta)$, $\tan(\alpha - \beta)$ con $\tan \alpha = -\frac{4}{3}$, $\alpha$ en $Q_2$, $\cos \beta = \frac{1}{2}$, $\beta$ en $Q_1$. Paso 1: Hallar $\sin \alpha$ y $\cos \alpha$ usando $\tan \alpha = \frac{\sin \alpha}{\cos \alpha} = -\frac{4}{3}$ y $\alpha$ en $Q_2$ (donde $\sin \alpha > 0$, $\cos \alpha < 0$). Sea $\cos \alpha = -3k$, $\sin \alpha = 4k$ para $k > 0$. Usando $\sin^2 \alpha + \cos^2 \alpha = 1$: $$(-3k)^2 + (4k)^2 = 9k^2 + 16k^2 = 25k^2 = 1 \Rightarrow k = \frac{1}{5}$$ Entonces: $$\cos \alpha = -\frac{3}{5}, \quad \sin \alpha = \frac{4}{5}$$ Paso 2: Hallar $\sin \beta$ usando $\cos^2 \beta + \sin^2 \beta = 1$ y $\beta$ en $Q_1$ (ambos positivos): $$\sin \beta = \sqrt{1 - \left(\frac{1}{2}\right)^2} = \sqrt{1 - \frac{1}{4}} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2}$$ Paso 3: Calcular $\sin(\alpha + \beta)$: $$\sin(\alpha + \beta) = \sin \alpha \cos \beta + \cos \alpha \sin \beta = \frac{4}{5} \cdot \frac{1}{2} + \left(-\frac{3}{5}\right) \cdot \frac{\sqrt{3}}{2} = \frac{2}{5} - \frac{3\sqrt{3}}{10} = \frac{4}{10} - \frac{3\sqrt{3}}{10} = \frac{4 - 3\sqrt{3}}{10}$$ Paso 4: Calcular $\cos(\alpha + \beta)$: $$\cos(\alpha + \beta) = \cos \alpha \cos \beta - \sin \alpha \sin \beta = \left(-\frac{3}{5}\right) \cdot \frac{1}{2} - \frac{4}{5} \cdot \frac{\sqrt{3}}{2} = -\frac{3}{10} - \frac{4\sqrt{3}}{10} = -\frac{3 + 4\sqrt{3}}{10}$$ Paso 5: Calcular $\tan \alpha = -\frac{4}{3}$ (dado) Paso 6: Calcular $\tan \beta = \frac{\sin \beta}{\cos \beta} = \frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}} = \sqrt{3}$ Paso 7: Calcular $\tan(\alpha - \beta)$: $$\tan(\alpha - \beta) = \frac{-\frac{4}{3} - \sqrt{3}}{1 + \left(-\frac{4}{3}\right) \cdot \sqrt{3}} = \frac{-\frac{4}{3} - \sqrt{3}}{1 - \frac{4\sqrt{3}}{3}} = \frac{-\frac{4}{3} - \sqrt{3}}{\frac{3 - 4\sqrt{3}}{3}} = \left(-\frac{4}{3} - \sqrt{3}\right) \cdot \frac{3}{3 - 4\sqrt{3}}$$ Multiplicamos numerador y denominador por el conjugado $3 + 4\sqrt{3}$: $$\tan(\alpha - \beta) = \frac{\left(-4 - 3\sqrt{3}\right)(3 + 4\sqrt{3})}{(3 - 4\sqrt{3})(3 + 4\sqrt{3})} = \frac{-12 - 16\sqrt{3} - 9\sqrt{3} - 12 \cdot 3}{9 - 48} = \frac{-12 - 25\sqrt{3} - 36}{-39} = \frac{-48 - 25\sqrt{3}}{-39} = \frac{48 + 25\sqrt{3}}{39}$$ --- 3. Problema: Determinar $\sin(\alpha + \beta)$, $\cos(\alpha + \beta)$, $\tan(\alpha - \beta)$ con $\sin \alpha = \frac{5}{13}$, $\alpha$ en $Q_4$, $\tan \beta = -\sqrt{3}$, $\beta$ en $Q_2$. Paso 1: Hallar $\cos \alpha$ usando $\sin^2 \alpha + \cos^2 \alpha = 1$ y $\alpha$ en $Q_4$ (donde $\cos \alpha > 0$): $$\cos \alpha = \sqrt{1 - \left(\frac{5}{13}\right)^2} = \sqrt{1 - \frac{25}{169}} = \sqrt{\frac{144}{169}} = \frac{12}{13}$$ Paso 2: Hallar $\sin \beta$ y $\cos \beta$ usando $\tan \beta = \frac{\sin \beta}{\cos \beta} = -\sqrt{3}$ y $\beta$ en $Q_2$ ($\sin \beta > 0$, $\cos \beta < 0$). Sea $\sin \beta = y$, $\cos \beta = x$, con $\frac{y}{x} = -\sqrt{3}$. Tomamos $x = -k$, $y = \sqrt{3}k$ para $k > 0$. Usando $x^2 + y^2 = 1$: $$(-k)^2 + (\sqrt{3}k)^2 = k^2 + 3k^2 = 4k^2 = 1 \Rightarrow k = \frac{1}{2}$$ Entonces: $$\cos \beta = -\frac{1}{2}, \quad \sin \beta = \frac{\sqrt{3}}{2}$$ Paso 3: Calcular $\sin(\alpha + \beta)$: $$\sin(\alpha + \beta) = \sin \alpha \cos \beta + \cos \alpha \sin \beta = \frac{5}{13} \cdot \left(-\frac{1}{2}\right) + \frac{12}{13} \cdot \frac{\sqrt{3}}{2} = -\frac{5}{26} + \frac{12\sqrt{3}}{26} = \frac{-5 + 12\sqrt{3}}{26}$$ Paso 4: Calcular $\cos(\alpha + \beta)$: $$\cos(\alpha + \beta) = \cos \alpha \cos \beta - \sin \alpha \sin \beta = \frac{12}{13} \cdot \left(-\frac{1}{2}\right) - \frac{5}{13} \cdot \frac{\sqrt{3}}{2} = -\frac{12}{26} - \frac{5\sqrt{3}}{26} = -\frac{12 + 5\sqrt{3}}{26}$$ Paso 5: Calcular $\tan \alpha = \frac{\sin \alpha}{\cos \alpha} = \frac{5/13}{12/13} = \frac{5}{12}$ Paso 6: Calcular $\tan(\alpha - \beta)$: $$\tan(\alpha - \beta) = \frac{\frac{5}{12} - (-\sqrt{3})}{1 + \frac{5}{12} \cdot (-\sqrt{3})} = \frac{\frac{5}{12} + \sqrt{3}}{1 - \frac{5\sqrt{3}}{12}} = \frac{\frac{5}{12} + \sqrt{3}}{\frac{12 - 5\sqrt{3}}{12}} = \left(\frac{5}{12} + \sqrt{3}\right) \cdot \frac{12}{12 - 5\sqrt{3}}$$ Multiplicamos numerador y denominador por el conjugado $12 + 5\sqrt{3}$: $$\tan(\alpha - \beta) = \frac{(5 + 12\sqrt{3})(12 + 5\sqrt{3})}{(12 - 5\sqrt{3})(12 + 5\sqrt{3})} = \frac{60 + 25\sqrt{3} + 144\sqrt{3} + 60 \cdot 3}{144 - 75} = \frac{60 + 169\sqrt{3} + 180}{69} = \frac{240 + 169\sqrt{3}}{69}$$ --- 4. Verificar identidad: $\sin\left(\frac{\pi}{2} + \theta\right) = \cos \theta$ Usando fórmula de suma: $$\sin\left(\frac{\pi}{2} + \theta\right) = \sin \frac{\pi}{2} \cos \theta + \cos \frac{\pi}{2} \sin \theta = 1 \cdot \cos \theta + 0 = \cos \theta$$ --- 5. Verificar identidad: $\cos\left(\frac{\pi}{2} + \theta\right) = -\sin \theta$ Usando fórmula de suma: $$\cos\left(\frac{\pi}{2} + \theta\right) = \cos \frac{\pi}{2} \cos \theta - \sin \frac{\pi}{2} \sin \theta = 0 - 1 \cdot \sin \theta = -\sin \theta$$ --- 6. Verificar identidad: $\tan(\pi - \theta) = -\tan \theta$ Usando fórmula de tangente de diferencia: $$\tan(\pi - \theta) = \frac{\tan \pi - \tan \theta}{1 + \tan \pi \tan \theta} = \frac{0 - \tan \theta}{1 + 0} = -\tan \theta$$ --- 7. Verificar identidad: $\cos\left(\frac{3\pi}{2} + \theta\right) = \sin \theta$ Usando fórmula de suma: $$\cos\left(\frac{3\pi}{2} + \theta\right) = \cos \frac{3\pi}{2} \cos \theta - \sin \frac{3\pi}{2} \sin \theta = 0 - (-1) \sin \theta = \sin \theta$$ --- 8. Verificar identidad: $\sin(\alpha + \beta) - \sin(\alpha - \beta) = 2 \sin \alpha \cos \beta$ Usando fórmulas de suma y resta: $$\sin(\alpha + \beta) = \sin \alpha \cos \beta + \cos \alpha \sin \beta$$ $$\sin(\alpha - \beta) = \sin \alpha \cos \beta - \cos \alpha \sin \beta$$ Restando: $$\sin(\alpha + \beta) - \sin(\alpha - \beta) = (\sin \alpha \cos \beta + \cos \alpha \sin \beta) - (\sin \alpha \cos \beta - \cos \alpha \sin \beta) = 2 \cos \alpha \sin \beta$$ La identidad dada es incorrecta, la correcta es: $$\sin(\alpha + \beta) - \sin(\alpha - \beta) = 2 \cos \alpha \sin \beta$$ --- 9. Verificar identidad: $\cot(\alpha + \beta) = \frac{\cot \alpha \cot \beta - 1}{\cot \alpha + \cot \beta}$ Usando $\cot x = \frac{\cos x}{\sin x}$ y fórmula de tangente: $$\cot(\alpha + \beta) = \frac{1}{\tan(\alpha + \beta)} = \frac{1}{\frac{\tan \alpha + \tan \beta}{1 - \tan \alpha \tan \beta}} = \frac{1 - \tan \alpha \tan \beta}{\tan \alpha + \tan \beta}$$ Reescribiendo en términos de cotangentes: $$\tan \alpha = \frac{1}{\cot \alpha}, \quad \tan \beta = \frac{1}{\cot \beta}$$ Entonces: $$\cot(\alpha + \beta) = \frac{1 - \frac{1}{\cot \alpha \cot \beta}}{\frac{1}{\cot \alpha} + \frac{1}{\cot \beta}} = \frac{\frac{\cot \alpha \cot \beta - 1}{\cot \alpha \cot \beta}}{\frac{\cot \beta + \cot \alpha}{\cot \alpha \cot \beta}} = \frac{\cot \alpha \cot \beta - 1}{\cot \alpha + \cot \beta}$$ --- 10. Verificar identidad: $\sec(\alpha + \beta) = \frac{\csc \alpha \csc \beta}{\cot \alpha \cot \beta - 1}$ Usando $\sec x = \frac{1}{\cos x}$, $\csc x = \frac{1}{\sin x}$, $\cot x = \frac{\cos x}{\sin x}$ y fórmula de coseno: $$\sec(\alpha + \beta) = \frac{1}{\cos(\alpha + \beta)} = \frac{1}{\cos \alpha \cos \beta - \sin \alpha \sin \beta}$$ Multiplicamos numerador y denominador por $\frac{1}{\sin \alpha \sin \beta}$: $$\sec(\alpha + \beta) = \frac{\frac{1}{\sin \alpha \sin \beta}}{\frac{\cos \alpha \cos \beta}{\sin \alpha \sin \beta} - 1} = \frac{\csc \alpha \csc \beta}{\cot \alpha \cot \beta - 1}$$ --- Respuesta final: 1) $\sin(\alpha + \beta) = \frac{2\sqrt{5}}{25}$, $\cos(\alpha + \beta) = \frac{11\sqrt{5}}{25}$, $\tan(\alpha - \beta) = 2$ 2) $\sin(\alpha + \beta) = \frac{4 - 3\sqrt{3}}{10}$, $\cos(\alpha + \beta) = -\frac{3 + 4\sqrt{3}}{10}$, $\tan(\alpha - \beta) = \frac{48 + 25\sqrt{3}}{39}$ 3) $\sin(\alpha + \beta) = \frac{-5 + 12\sqrt{3}}{26}$, $\cos(\alpha + \beta) = -\frac{12 + 5\sqrt{3}}{26}$, $\tan(\alpha - \beta) = \frac{240 + 169\sqrt{3}}{69}$ 4) $\sin\left(\frac{\pi}{2} + \theta\right) = \cos \theta$ 5) $\cos\left(\frac{\pi}{2} + \theta\right) = -\sin \theta$ 6) $\tan(\pi - \theta) = -\tan \theta$ 7) $\cos\left(\frac{3\pi}{2} + \theta\right) = \sin \theta$ 8) $\sin(\alpha + \beta) - \sin(\alpha - \beta) = 2 \cos \alpha \sin \beta$ 9) $\cot(\alpha + \beta) = \frac{\cot \alpha \cot \beta - 1}{\cot \alpha + \cot \beta}$ 10) $\sec(\alpha + \beta) = \frac{\csc \alpha \csc \beta}{\cot \alpha \cot \beta - 1}$