Question

hallar el valor de todas lass funciones trigonometricas de ...

Answer 1

Respuesta:

Espero te sirva

Answer 2

Respuesta:

Explicación paso a paso:

a) θ ∈ I cuadrante ,  x, y positivos  

$sin\ \theta = \frac{y}{R} = \frac{cateto\ opuesto\ al\ angulo\ \theta }{hipotenusa}$

$cos\ \theta = \frac{y}{R} = \frac{cateto\ adyacente\ al\ angulo\ \theta }{hipotenusa}$

$sin\ \theta =\frac{1}{3}$     ;    $cos\ \theta =\frac{2\sqrt{3} }{3}$    

$tan\ \theta = \frac{y}{x} = \frac{1}{2\sqrt{3} } *\frac{\sqrt{3} }{\sqrt{3} } = \frac{\sqrt{3} }{6}$

$cot\ \theta = \frac{x}{y} = \frac{2\sqrt{3} }{1 }= 2\sqrt{3}$

$sec\ \theta = \frac{R}{x} = \frac{3}{2\sqrt{3} } *\frac{\sqrt{3} }{\sqrt{3} } = \frac{\sqrt{3} }{2}$

$csc\ \theta = \frac{R}{y} = \frac{3}{1 } = 3$

b) θ ∈ I cuadrante ,  x, y positivos

x = 2  ;  y = $\sqrt{21}$  ;  R = 5

$sen\ \theta = \frac{\sqrt{21} }{5}$      ;     $cot\ \theta = \frac{2}{\sqrt{21} } = \frac{2\sqrt{21} }{21}$

$sec\ \theta = \frac{5 }{2 }$           ;    $csc\ \theta = \frac{5 }{\sqrt{21} } = \frac{5\sqrt{21} }{21}$

c) θ ∈ IV cuadrante ,  x(+) ; y(-)

x = 2$\sqrt{2}$  ;   y = - 1  ;   R = 5

$sen\ \theta = \frac{\sqrt{21} }{5}$      ;     $cot\ \theta = \frac{2}{\sqrt{21} } = \frac{2\sqrt{21} }{21}$

$sec\ \theta = \frac{5 }{2 }$           ;    $csc\ \theta = \frac{5 }{\sqrt{21} } = \frac{5\sqrt{21} }{21}$

d) θ ∈ I cuadrante ,  x, y positivos

x = $\sqrt{3}$  ;  y = 1  ;  R = 2

$sen\ \theta = \frac{1 }{2}$                ;     $tan\ \theta = \frac{1}{\sqrt{3} } = \frac{\sqrt{3} }{3}$

$sec\ \theta = \frac{2 }{\sqrt{3} }= \frac{2\sqrt{3} }{3}$    ;    $csc\ \theta = \frac{2}{1} = 2$

e) θ ∈ II cuadrante ,  x(-) , y(+)

x = - 1 ;  y = 1  ;  R = $\sqrt{2}$

$tan\ \theta = \frac{1 }{-1} = -1$       ;     $cot\ \theta = \frac{-1}{1} = -1$

$sec\ \theta = \frac{\sqrt{2} }{-1 }= -\sqrt{2}$    ;    $sec\ \theta = \frac{\sqrt{2} }{1 }= \sqrt{2}$

f) θ ∈ IV cuadrante ,  x(+), y(-)

x = $\sqrt{2}$  ;  y =  - 1  ;  R = $\sqrt{3}$

$sen\ \theta = \frac{-1 }{\sqrt{3} }=-\frac{\sqrt{3} }{3}$     ;    $cos\ \theta = \frac{\sqrt{2} }{\sqrt{3} } = \frac{\sqrt{6} }{3}$

$sec\ \theta = \frac{\sqrt{3} }{\sqrt{2} }= \frac{\sqrt{6} }{2}$        ;    $csc\ \theta = \frac{\sqrt{3} }{-1} = -\sqrt{3}$