p(1, 0) corresponds to ( , ) on the cosine graph.\ndone\np(0, 1) corresponds to ( , ) on the cosine…

Explanation:

Step1: Recall cosine - angle relationship

For a point $P(x,y)$ on the unit - circle, $y = \cos\theta$. Also, $\theta$ is the angle measured counter - clockwise from the positive $x$ - axis.

Step2: Analyze $P(1,0)$

When $P(1,0)$, the angle $\theta = 0^{\circ}$ (or $0$ radians) and $\cos\theta=\cos(0)=1$. So the point on the cosine graph is $(0,1)$.

Step3: Analyze $P(0,1)$

When $P(0,1)$, the angle $\theta = 90^{\circ}$ (or $\frac{\pi}{2}$ radians) and $\cos\theta=\cos(90^{\circ}) = 0$. So the point on the cosine graph is $(90,0)$.

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

When $P(-1,0)$, the angle $\theta = 180^{\circ}$ (or $\pi$ radians) and $\cos\theta=\cos(180^{\circ})=-1$. So the point on the cosine graph is $(180, - 1)$.

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

When $P(0,-1)$, the angle $\theta = 270^{\circ}$ (or $\frac{3\pi}{2}$ radians) and $\cos\theta=\cos(270^{\circ}) = 0$. So the point on the cosine graph is $(270,0)$.

Answer:

$P(1,0)$ corresponds to $(0,1)$ on the cosine graph. $P(0,1)$ corresponds to $(90,0)$ on the cosine graph. $P(-1,0)$ corresponds to $(180,-1)$ on the cosine graph. $P(0,-1)$ corresponds to $(270,0)$ on the cosine graph.