Final answer:
To find sin θ and cos θ in different quadrants when given tan θ, we can use the properties of right triangles and the trigonometric functions. In quadrant IV with tan θ = -1/5, sin θ = (-1)/√26 and cos θ = 5/√26. In quadrant IV with tan θ = -2/3, sin θ = (-2)/√13 and cos θ = 3/√13. In quadrant III with tan θ = 2, sin θ = 2/√5 and cos θ = (-1)/√5.
Step-by-step explanation:
To find sin θ and cos θ when tan θ = -1/5 and the terminal side lies in quadrant IV, we can use the properties of the trigonometric functions in the right triangle. Since tan θ = -1/5, we can let the opposite side of the triangle be -1 and the adjacent side be 5. Using the Pythagorean theorem, we find that the hypotenuse is √(5^2 + (-1)^2) = √26. Therefore, sin θ = (-1)/√26 and cos θ = 5/√26.
To find sin θ and cos θ when tan θ = -2/3 and the terminal side lies in quadrant IV, we can use similar steps. Let the opposite side be -2 and the adjacent side be 3. Using the Pythagorean theorem, we find the hypotenuse is √(3^2 + (-2^2)) = √13. Therefore, sin θ = (-2)/√13 and cos θ = 3/√13.
To find sin θ and cos θ when tan θ = 2 and the terminal side lies in quadrant III, we can let the opposite side be 2 and the adjacent side be -1. Using the Pythagorean theorem, we find the hypotenuse is √((-1)^2 + 2^2) = √5. Therefore, sin θ = 2/√5 and cos θ = (-1)/√5.