csc thea= -2, quadrant 4. what are the six trig functions?

in QIV, x>0 y<0

csc = -2 so
sin = -1/2
cos = √3/2
...

To find the values of the six trigonometric functions at an angle in a specific quadrant, you first need to determine the values of sine (sin) and cosine (cos) for that angle in the given quadrant.

In this case, you're given that csc θ (cosecant) is -2 and that the angle θ is in quadrant 4. Since csc is the reciprocal of sin, we can find sin θ by taking the reciprocal of -2, which gives us sin θ = -1/2.

To determine the value of cos θ, we can use the Pythagorean identity: sin^2 θ + cos^2 θ = 1. Plugging in the value we found for sin θ, we have (-1/2)^2 + cos^2 θ = 1. Simplifying this equation, we get 1/4 + cos^2 θ = 1. Subtracting 1/4 from both sides gives cos^2 θ = 3/4. Taking the square root of both sides, we find cos θ = ±√(3/4). However, since θ is in quadrant 4 (where cosine is positive), we take the positive value, so cos θ = √(3/4) = √3/2.

Now that we have sin θ = -1/2 and cos θ = √3/2, we can calculate the remaining four trigonometric functions:

1. Tangent (tan θ) is the ratio of sin θ to cos θ:
tan θ = sin θ / cos θ = (-1/2) / (√3/2) = -1/√3 = -√3/3

2. Cotangent (cot θ) is the reciprocal of tangent:
cot θ = 1 / tan θ = 1 / (-√3/3) = -√3

3. Secant (sec θ) is the reciprocal of cosine:
sec θ = 1 / cos θ = 1 / (√3/2) = 2 / √3 = 2√3/3

4. Cosecant (csc θ) is the reciprocal of sine:
csc θ = 1 / sin θ = 1 / (-1/2) = -2

Therefore, the six trigonometric functions for an angle θ in quadrant 4, given csc θ = -2, are:
sin θ = -1/2
cos θ = √3/2
tan θ = -√3/3
cot θ = -√3
sec θ = 2√3/3
csc θ = -2