If sin(theta) = -4/5.

Question

Then find Cos(theta) and tan of (theta)?  

 

 

Answer 1

a) cosθ

We have sin²θ + cos²θ = 1

Substitute sinθ value.

(-4/5)² + cos²θ = 1

16/25 + cos²θ = 1

cos²θ = 1 - 16/25

cos²θ = (25 - 16) / 25

cos²θ = 9 / 25

cosθ = √(9 / 25)

cosθ = √(9) / √(25)

cosθ = 3 / 5

b) tanθ

tanθ = sinθ / cosθ

(-4 / 5) / (3 / 5) = - 4 / 3

Therefore tanθ = - 4/3

Therefore cosθ = 3 / 5 and tan(t) = -4 / 3.

Answer 2

The trigonometric function sin θ = - 4/5.

Using the Pythagorean identity sin² θ + cos² θ = 1, obtain

(- 4/5)² + cos² θ = 1

cos² θ = 1 - 16/25 = 9/25

The function sin(θ) is negative in third and fourth quadrant, so the angle θ lies in either third or fourth quadrant.

In third Quadrant ( π ≤ θ ≤ 3π/2 ), function cos(θ) is negative, use the negative root obtain

cos θ = - √9/√25 = - 3/5.

Use Quotient Identities tan(θ) = sin(θ)/cos(θ) = (- 4/5)/(- 3/5) = 4/3.

In fourth Quadrant ( 3π/2 ≤ θ ≤ 2π ), function cos(θ) is positive, use the positive root obtain

cos θ = √9/√25 = 3/5.

Use Quotient Identities tan(θ) = sin(θ)/cos(θ) = (- 4/5)/(3/5) = - 4/3.

In Quadrant III, cos θ = - 3/5 and tan(θ) = 4/3.

In Quadrant IV, cos θ = 3/5 and tan(θ) = - 4/3.