essential question what are the characteristics of the graphs of the sine and cosine functions?\n1…

Explanation:

Step1: Recall sine - function values

Use the unit - circle definition of the sine function $\sin x = y$ - coordinate of the point on the unit circle corresponding to the angle $x$.

Step2: Calculate values for $x=-2\pi$

Since $\sin(-2\pi)=0$ (because the angle $- 2\pi$ corresponds to the point $(1,0)$ on the unit circle).

Step3: Calculate values for $x =-\frac{7\pi}{4}$

$\sin(-\frac{7\pi}{4})=\frac{\sqrt{2}}{2}$, as the angle $-\frac{7\pi}{4}$ is equivalent to $\frac{\pi}{4}$ in the positive direction and $\sin\frac{\pi}{4}=\frac{\sqrt{2}}{2}$.

Step4: Calculate values for $x =-\frac{3\pi}{2}$

$\sin(-\frac{3\pi}{2}) = 1$, because the angle $-\frac{3\pi}{2}$ corresponds to the point $(0,1)$ on the unit circle.

Step5: Calculate values for $x =-\frac{5\pi}{4}$

$\sin(-\frac{5\pi}{4})=\frac{\sqrt{2}}{2}$, since $-\frac{5\pi}{4}$ is equivalent to $\frac{3\pi}{4}$ in the positive direction and $\sin\frac{3\pi}{4}=\frac{\sqrt{2}}{2}$.

Step6: Calculate values for $x =-\pi$

$\sin(-\pi)=0$, as the angle $-\pi$ corresponds to the point $(-1,0)$ on the unit circle.

Step7: Calculate values for $x =-\frac{3\pi}{4}$

$\sin(-\frac{3\pi}{4})=-\frac{\sqrt{2}}{2}$, because $-\frac{3\pi}{4}$ is equivalent to $\frac{5\pi}{4}$ in the positive direction and $\sin\frac{5\pi}{4}=-\frac{\sqrt{2}}{2}$.

Step8: Calculate values for $x =-\frac{\pi}{2}$

$\sin(-\frac{\pi}{2})=-1$, since the angle $-\frac{\pi}{2}$ corresponds to the point $(0, - 1)$ on the unit circle.

Step9: Calculate values for $x =-\frac{\pi}{4}$

$\sin(-\frac{\pi}{4})=-\frac{\sqrt{2}}{2}$, as $-\frac{\pi}{4}$ is equivalent to $\frac{7\pi}{4}$ in the positive direction and $\sin\frac{7\pi}{4}=-\frac{\sqrt{2}}{2}$.

Step10: Calculate values for $x = 0$

$\sin(0)=0$, corresponding to the point $(1,0)$ on the unit circle.

Step11: Calculate values for $x=\frac{\pi}{4}$

$\sin\frac{\pi}{4}=\frac{\sqrt{2}}{2}$.

Step12: Calculate values for $x=\frac{\pi}{2}$

$\sin\frac{\pi}{2}=1$.

Step13: Calculate values for $x=\frac{3\pi}{4}$

$\sin\frac{3\pi}{4}=\frac{\sqrt{2}}{2}$.

Step14: Calculate values for $x=\pi$

$\sin(\pi)=0$.

Step15: Calculate values for $x=\frac{5\pi}{4}$

$\sin\frac{5\pi}{4}=-\frac{\sqrt{2}}{2}$.

Step16: Calculate values for $x=\frac{3\pi}{2}$

$\sin\frac{3\pi}{2}=-1$.

Step17: Calculate values for $x=\frac{7\pi}{4}$

$\sin\frac{7\pi}{4}=-\frac{\sqrt{2}}{2}$.

Step18: Calculate values for $x = 2\pi$

$\sin(2\pi)=0$.

Step19: Calculate values for $x=\frac{9\pi}{4}$

$\sin\frac{9\pi}{4}=\frac{\sqrt{2}}{2}$, since $\frac{9\pi}{4}=2\pi+\frac{\pi}{4}$ and $\sin(2\pi +\alpha)=\sin\alpha$.

The completed table:

$x$	$-2\pi$	$-\frac{7\pi}{4}$	$-\frac{3\pi}{2}$	$-\frac{5\pi}{4}$	$-\pi$	$-\frac{3\pi}{4}$	$-\frac{\pi}{2}$	$-\frac{\pi}{4}$	$0$	$\frac{\pi}{4}$	$\frac{\pi}{2}$	$\frac{3\pi}{4}$	$\pi$	$\frac{5\pi}{4}$	$\frac{3\pi}{2}$	$\frac{7\pi}{4}$	$2\pi$	$\frac{9\pi}{4}$
$y = \sin x$	$0$	$\frac{\sqrt{2}}{2}$	$1$	$\frac{\sqrt{2}}{2}$	$0$	$-\frac{\sqrt{2}}{2}$	$-1$	$-\frac{\sqrt{2}}{2}$	$0$	$\frac{\sqrt{2}}{2}$	$1$	$\frac{\sqrt{2}}{2}$	$0$	$-\frac{\sqrt{2}}{2}$	$-1$	$-\frac{\sqrt{2}}{2}$	$0$	$\frac{\sqrt{2}}{2}$

For part (b), after plotting the points $(x,y)$ from the table on the $x - y$ coordinate plane, we can draw a smooth curve that oscillates between $y=-1$ and $y = 1$, has a period of $2\pi$, and passes through the origin. The curve is symmetric about the origin (since $y = \sin x$ is an odd function, $\sin(-x)=-\sin x$).

Answer:

The completed table is shown above and the graph of $y = \sin x$ is a smooth - oscillating curve between $y=-1$ and $y = 1$ with period $2\pi$ passing through the origin.