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

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

region b will make the inequalities y < 2x + 1 and y > -x - 2 true ⇒ answer a

step-by-step explanation:

* lets study the graph and the inequality to solve the question

- the inequality y > -x - 2

∵ the form of the inequality is y > mx + c , m is the slope of the line

  and c is the y-intercept (the line intersect y-axis at point (0 , c)

∴ c = -2 that means the line intersect the y-axis at point (0 , -2)

∵ the sign of the inequality is >

∴ we shad the part over the line

  region a and b

- the inequality y < 2x + 1

∵ the form of the inequality is y > mx + c , m is the slope of the line

  and c is the y-intercept (the line intersect y-axis at point (0 , c)

∴ c = 1 that means the line intersect the y-axis at point (0 , 1)

∵ the sign of the inequality is <

∴ we shad the part under the line

  region b and c

∵ the common region between the two lines is region b

∴ region b will make the inequalities y < 2x + 1 and y > -x - 2 true