Ii
MATU • Trigonometria
MATU_TRI_227
Guía de Ejercicios
Enunciado
Calcular $\sin \frac{\alpha}{2}, \cos \frac{\alpha}{2}, \tan \frac{\alpha}{2}$ si:
(a) $\cos \alpha = 0.8$ y $0 < \alpha < \frac{\pi}{2}$;
(b) $\tan \alpha = 3 \frac{3}{7}$ y $180^\circ < \alpha < 270^\circ$.
(a) $\cos \alpha = 0.8$ y $0 < \alpha < \frac{\pi}{2}$;
(b) $\tan \alpha = 3 \frac{3}{7}$ y $180^\circ < \alpha < 270^\circ$.
Solución Paso a Paso
Parte (a): $\cos \alpha = 4/5$ en I cuadrante. Entonces $\alpha/2$ también está en I cuadrante (signos $+$).
$$ \sin \frac{\alpha}{2} = \sqrt{\frac{1 - 0.8}{2}} = \sqrt{0.1} = \frac{1}{\sqrt{10}} $$
$$ \cos \frac{\alpha}{2} = \sqrt{\frac{1 + 0.8}{2}} = \sqrt{0.9} = \frac{3}{\sqrt{10}} $$
$$ \tan \frac{\alpha}{2} = \frac{1/ \sqrt{10}}{3/ \sqrt{10}} = \frac{1}{3} $$
Parte (b): $\tan \alpha = 24/7$. En III cuadrante, $\sin \alpha = -24/25$ y $\cos \alpha = -7/25$.
Si $180^\circ < \alpha < 270^\circ$, entonces $90^\circ < \alpha/2 < 135^\circ$ (II cuadrante).
$\sin$ es $(+)$, $\cos$ es $(-)$, $\tan$ es $(-)$.
$$ \sin \frac{\alpha}{2} = \sqrt{\frac{1 - (-7/25)}{2}} = \sqrt{\frac{32}{50}} = \frac{4}{5} $$
$$ \cos \frac{\alpha}{2} = -\sqrt{\frac{1 + (-7/25)}{2}} = -\sqrt{\frac{18}{50}} = -\frac{3}{5} $$
$$ \tan \frac{\alpha}{2} = \frac{4/5}{-3/5} = -\frac{4}{3} $$
$$ \sin \frac{\alpha}{2} = \sqrt{\frac{1 - 0.8}{2}} = \sqrt{0.1} = \frac{1}{\sqrt{10}} $$
$$ \cos \frac{\alpha}{2} = \sqrt{\frac{1 + 0.8}{2}} = \sqrt{0.9} = \frac{3}{\sqrt{10}} $$
$$ \tan \frac{\alpha}{2} = \frac{1/ \sqrt{10}}{3/ \sqrt{10}} = \frac{1}{3} $$
Parte (b): $\tan \alpha = 24/7$. En III cuadrante, $\sin \alpha = -24/25$ y $\cos \alpha = -7/25$.
Si $180^\circ < \alpha < 270^\circ$, entonces $90^\circ < \alpha/2 < 135^\circ$ (II cuadrante).
$\sin$ es $(+)$, $\cos$ es $(-)$, $\tan$ es $(-)$.
$$ \sin \frac{\alpha}{2} = \sqrt{\frac{1 - (-7/25)}{2}} = \sqrt{\frac{32}{50}} = \frac{4}{5} $$
$$ \cos \frac{\alpha}{2} = -\sqrt{\frac{1 + (-7/25)}{2}} = -\sqrt{\frac{18}{50}} = -\frac{3}{5} $$
$$ \tan \frac{\alpha}{2} = \frac{4/5}{-3/5} = -\frac{4}{3} $$