p(1, 0) corresponds to ( ) on the cosine graph. done p(0, 1) corresponds to ( ) on the cosine graph. done…

Answer:

1. For (P(1,0)): ((0,1))
2. For (P(0,1)): ((90^{\circ},0))
3. For (P(-1,0)): ((180^{\circ}, - 1))
4. For (P(0,-1)): ((270^{\circ},0))

Explanation:

Step1: Recall cosine - unit circle relationship

The cosine function (y = \cos\theta) where (\theta) is the angle measured counter - clockwise from the positive (x) - axis on the unit circle. The (x) - coordinate of a point on the unit circle is (\cos\theta) and the (y) - coordinate is (\sin\theta).

Step2: Analyze (P(1,0))

When (x = 1) and (y = 0) on the unit circle, (\theta=0^{\circ}). And (\cos(0^{\circ}) = 1), so on the (y = \cos\theta) graph, the point is ((0,1)).

Step3: Analyze (P(0,1))

When (x = 0) and (y = 1) on the unit circle, (\theta = 90^{\circ}). And (\cos(90^{\circ})=0), so on the (y=\cos\theta) graph, the point is ((90^{\circ},0)).

Step4: Analyze (P(-1,0))

When (x=-1) and (y = 0) on the unit circle, (\theta = 180^{\circ}). And (\cos(180^{\circ})=-1), so on the (y=\cos\theta) graph, the point is ((180^{\circ}, - 1)).

Step5: Analyze (P(0,-1))

When (x = 0) and (y=-1) on the unit circle, (\theta = 270^{\circ}). And (\cos(270^{\circ}) = 0), so on the (y=\cos\theta) graph, the point is ((270^{\circ},0)).