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Sabiendo que la tangente son 4/3 dime el cos y sen estando el angulo en el tercer cuadrante de la circunferencia

Sagot :

Answer:

sen α  =  - 4/5

cos α  =  - 3/5

Step-by-step explanation:

tang α  = 4/3

De la relación fundamental de la trigonometría sabemos:

sen² α   +  cos²α  = 1

Dividiendo la ecuación anterior por cos² α

sen²α/cos²α   +  cos²α / cos²α   =  1 / cos² α

tang²α  +  1   = 1/ cos² α

(4/3)²  +  1  =  1/ cos² α    ⇒    16/9   +  1  = 1/ cos² α

1/ cos² α    =  ( 16  +  9  ) / 9

1/ cos² α    =   25/9

cos² α   = 9/25

cos α    =  3/5

y para calcular el sen α  

sen² α   =  1  - cos²α

sen² α   =  1  - 9/25

sen² α   =  ( 25  -  9 )/ 25

sen² α   =  16/25

sen α  = 4 / 5

Ahora bien el angulo α  está en el tercer cuadrante, en ese cuadrante tanto el seno como el coseno son negativos.

Respuesta:

sen α  =  - 4/5

cos α  =  - 3/5

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