Answer:-(√3)/2, (√2)/2, -√3, and undefined
Step-by-step explanation:
There are two ways you can solve this. One is with the Pythagorean identity:
sin²θ + cos²θ = 1
The other way is by knowing your unit circle.
1. From the unit circle, we know that cos θ = 1/2 at θ = π/3 and θ = 5π/3. Since θ is in Quadrant IV, then θ = 5π/3. sin (5π/3) = -(√3)/2.
We can check our answer using the Pythagorean identity:
sin²θ + cos²θ = 1
sin²θ + (1/2)² = 1
sin²θ + 1/4 = 1
sin²θ = 3/4
sin θ = ±(√3)/2
Since sine is negative in Quadrant IV, sin θ = -(√3)/2.
We can repeat these steps for the other questions.
2. sin θ = (√2)/2 at θ = π/4 and θ = 3π/4. Since θ is in Quadrant I, θ = π/4. Therefore, cos θ = (√2)/2.
3. cos θ = -1/2 at θ = 2π/3 and θ = 4π/3. Since θ is in Quadrant II, θ = 2π/3. Therefore, sin θ = (√3)/2, and tan θ = sin θ / cos θ = -√3.
4. sin θ = -1 at θ = 3π/2. Therefore, cos θ = 0. tan θ = sin θ / cos θ, so tan θ is undefined.
