Question

Empareje cada expresion de la izquierda con el valor correcto de la derecha.
a) tan 5π\ 6 e)√3
b)cot 5π\ 6 f) - √3
c) tan 7π / 6 g) 1 /√3
d) cot 7π/ 6 h) -1 / √3

Answer 1

Utilizando identidades trigonometricas y el valor de los angulos notables:

$tan( \frac{5\pi}{6})= \frac{sen(\frac{5\pi}{6})}{cos(\frac{5\pi}{6})}$

Pero:

$sen(\frac{5\pi}{6})=sen(\frac{2\pi}{6}+\frac{3\pi}{6}) \\ sen(\frac{5\pi}{6})=sen(\frac{\pi}{3}+\frac{\pi}{2}) \\ sen(\frac{5\pi}{6})=sen(\frac{\pi}{3})cos(\frac{\pi}{2})+sen(\frac{\pi}{2})cos(\frac{\pi}{3}) \\ sen(\frac{5\pi}{6})= \frac{\sqrt{3}}{2} *(0) + (1)*( \frac{1}{2}) \\ sen(\frac{5\pi}{6})=\frac{1}{2}$

Y:

$cos(\frac{5\pi}{6})=sen(\frac{2\pi}{6}+\frac{3\pi}{6}) \\ cos(\frac{5\pi}{6})=cos(\frac{\pi}{3}+\frac{\pi}{2}) \\ cos(\frac{5\pi}{6})=cos(\frac{\pi}{3})cos(\frac{\pi}{2})-sen(\frac{\pi}{2})sen(\frac{\pi}{3}) \\ cos(\frac{5\pi}{6})= \frac{1}{2} *(0) - (1)*( \frac{\sqrt{3}}{2}) \\ cos(\frac{5\pi}{6})=-\frac{\sqrt{3}}{2}$

Entonces:

$tan( \frac{5\pi}{6})= \frac{sen(\frac{5\pi}{6})}{cos(\frac{5\pi}{6})} \\ tan( \frac{5\pi}{6})= \frac{ \frac{1}{2} }{ -\frac{ \sqrt{3} }{2} } \\ tan( \frac{5\pi}{6})= -\frac{2}{2\sqrt{3}} \\ tan( \frac{5\pi}{6})=-\frac{1}{\sqrt{3}}$

Ahora:

$cot(\frac{5\pi}{6})= \frac{1}{tan( \frac{5\pi}{6})} \\ cot(\frac{5\pi}{6})= \frac{1}{-\frac{1}{\sqrt{3}}} \\ cot(\frac{5\pi}{6})=-\sqrt{3}$

Haciendo un analisis similar a:

$tan( \frac{7\pi}{6})= \frac{sen(\frac{7\pi}{6})}{cos(\frac{7\pi}{6})}$

Pero:

$sen(\frac{7\pi}{6})=sen(\frac{4\pi}{6}+\frac{3\pi}{6}) \\ sen(\frac{7\pi}{6})=sen(\frac{2\pi}{3}+\frac{\pi}{2}) \\ sen(\frac{7\pi}{6})=sen(\frac{2\pi}{3})cos(\frac{\pi}{2})+sen(\frac{\pi}{2})cos(\frac{2\pi}{3}) \\ sen(\frac{7\pi}{6})=(\frac{\sqrt{3}}{2}) *(0) + (1)*( -\frac{1}{2}) \\ sen(\frac{7\pi}{6})=-\frac{1}{2}$

Y:
$cos(\frac{7\pi}{6})=cos(\frac{4\pi}{6}+\frac{3\pi}{6}) \\ cos(\frac{7\pi}{6})=cos(\frac{2\pi}{3}+\frac{\pi}{2}) \\ cos(\frac{7\pi}{6})=cos(\frac{2\pi}{3})cos(\frac{\pi}{2})-sen(\frac{\pi}{2})sen(\frac{2\pi}{3}) \\ cos(\frac{7\pi}{6})=(-\frac{1}{2}) *(0) - (1)*( \frac{\sqrt{3}}{2}) \\ cos(\frac{7\pi}{6})=-\frac{\sqrt{3}}{2}$

Entonces: 

$tan( \frac{7\pi}{6})= \frac{sen(\frac{7\pi}{6})}{cos(\frac{7\pi}{6})} \\ tan( \frac{7\pi}{6})= \frac{- \frac{1}{2} }{-\frac{\sqrt{3}}{2}} \\ tan( \frac{7\pi}{6})= \frac{2}{2\sqrt{3}} \\ tan( \frac{7\pi}{6})= \frac{1}{\sqrt{3}}$

Ahora:

$cot(\frac{7\pi}{6})= \frac{1}{tan( \frac{7\pi}{6})} \\ cot(\frac{7\pi}{6})= \frac{1}{\frac{1}{\sqrt{3}}} \\ cot(\frac{7\pi}{6})= \sqrt{3}$