If sin(θ) = -3/5, and θ is in quadrant III, then find (a) cos (θ) = (b) tan (θ) = (c) sec (θ) = (d) csc (θ) = (e) cot (θ) =

Question

If sin(θ) = -3/5, and θ is in quadrant III, then find

(a) cos (θ) =

(b) tan (θ) =

(c) sec (θ) =

(d) csc (θ) =

(e) cot (θ) =

Answer 1

This triangle is a Pythagorean triple (3-4-5) if you did not do that you can do sin^-1(-3/5)=θ and then cos(θ)=-4/5

Now we know the sides, 3 is opposite, 4 is adjacent, 5 is hypotenuse (we know this because sin is opposite over hypotenuse), and because it is in Quadrant 3 x and y/ sin and cos are negative but tan is positive/

(a) cos (θ) = -4/5

(b) tan (θ) = 3/4

(c) sec (θ) = -5/4

(d) csc (θ) = -5/3

(e) cot (θ) = 4/3