1. find sinθ if cosθ=1/2 and θ terminates in quadrant iv.
2. find cosθ if sinθ=(√2)/2 and θ terminates in quadrant i.
3. find tanθ if cosθ=-1/2 and θ terminates in quadrant ii.
4. find tanθ if sinθ=-1 and 0≤θ< 2π radians.

-(√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.

logv3(x+2)=1

rewrite in exponential form:

logb(x) = y is equal to b^y =x

b = 3, y = 1

now solve for x:

3^1 = x+2)

x = 3-2

x = 1

see the picture for the graph.

i think the answer is seven, for if i'm being useless °ω°

step-by-step explanation: