Graphing the basic sine and cosine curves.

Question

1. Draw $x$ and $y$ axes, marking the $x$ axis with tick marks that are a multiple of $\pi$ and the $y$ axis with integers to fit the problem.

2. Determine all $x$-intercepts by solving an equation such as $x = \sin^{-1} 0$ or $x = \cos^{-1} 0$ for all $x$-values in the interval you're graphing.

3. Graph the intercepts.

4. Determine the $y$-intercept by letting $x = 0$ and solving for $y$ in the equation.

5. Graph the $y$-intercept.

6. Determine where the extreme values are by solving an equation such as $x = \sin^{-1} \pm 1$ or $x = \cos^{-1} \pm 1$ for all $x$-values in the interval you're graphing.

7. Graph those values.

8. Sketch in the curve.

Example problem:

Graph $y = \sin x$ between $x = 2\pi$ and $x = 4\pi$.

A. To find the $x$-intercepts, solve $x = \sin^{-1} 0$ to get $x = 2\pi$ or $3\pi$ or $4\pi$. There's no $y$-intercept here, because the $y$-axis isn't in that interval. The function is equal to $1$ at $x = \frac {5\pi}{2}$ and $-1$ at $x = \frac {7\pi}{2}$. The graph is shown in the figure.

My questions are this:

1. How does one come up with $2\pi$ or $3\pi$ or $4\pi$ from $x = \sin^{-1} 0$? As the equation equals $0$ and only computes between $1$ and $-1$.

2. I'm having trouble on how to compute the $y$-intercept. I'm going to say you just solve $y = \sin x$? In this case plug in $2\pi$ or $3\pi$ or $4\pi$?In the answer it says "there's no $y$-intercept here, because the $y$-axis isn't in that interval". Do they mean the range of $2\pi$ to $4\pi$, when they say interval?

3. Number $6$ in the $8$ points up top. When I determine extreme values do I just solve $y = \sin x$ and then plug the values into $x = \sin^{-1} \pm 1$ or $x = \cos^{-1} \pm 1$?

Answer 1

Let's use the definition that the sine of an angle in standard position (vertex at the origin, initial side on the positive $x$-axis) is the $y$-coordinate of the point where the terminal side of the angle intersects the unit circle.

Therefore, the graph of the sine function $f(x) = \sin x$ is obtained by tracing out the $y$-coordinates of the point where the terminal side of the angle intersects the unit circle as it is rotated around the circle.

Once a full revolution of the circle is completed, continuing to rotate the terminal side of the angle around the circle causes the values to repeat. Hence, the sine function is periodic. Since there are $2\pi$ radians in a full circle, $$f(x + 2\pi) = \sin(x + 2\pi) = \sin x = f(x)$$
Since there is no smaller positive value $p$ such that $f(x + p) = f(x)$ for each $p$ in its domain, the sine function has period $2\pi$.

You were asked to restrict the domain of the function to the interval $[2\pi, 4\pi]$, which yields the graph shown below.

Let's call the function in your example $g$. Then $g: [2\pi, 4\pi] \to [-1, 1]$ is the function defined by $g(x) = \sin x$.

Since $0$ is not in the domain of $g$, it does not have a $y$-intercept.

The sine function $f(x) = \sin x$ has an $x$-intercept at each point where the terminal side of the angle lies on the $x$-axis, which occurs when $x = n\pi, n \in \mathbb{Z}$. The three values where this occurs in the domain of $g$ are $2\pi$, $3\pi$, and $4\pi$.

The sine function $f(x) = \sin x$ reaches its maximum value of $1$ whenever the terminal side of the angle lies on the positive $y$-axis, which occurs when \begin{align*} x & = \arcsin (1) + 2n\pi, n \in \mathbb{Z}\\ & = \frac{\pi}{2} + 2n\pi, n \in \mathbb{Z} \end{align*} where the $2n\pi$ term is used to account for the periodicity of the sine function as all angles that differ by an integer multiple of $2\pi$ have the same sine value. The only such value in the domain of $g$ is $$\frac{\pi}{2} + 2\pi = \frac{5\pi}{2}$$

The sine function $f(x) = x$ reaches its minimum value of $-1$ whenever the terminal side of the angle lies on the negative $y$-axis, which occurs when \begin{align*} x & = \arcsin(-1) + 2n\pi, n \in \mathbb{Z}\\ & = -\frac{\pi}{2} + 2n\pi, n \in \mathbb{Z} \end{align*} The only such value in the domain of $g$ is $$x = -\frac{\pi}{2} + 4\pi = \frac{7\pi}{2}$$ Again, the $2n\pi$ term is used to account for the periodicity of the sine function.

I have addressed your second and third questions above. As for your first question, note that the inverse sine function $h: [-1, 1] \to [-\frac{\pi}{2}, \frac{\pi}{2}]$, denoted by $h(x) = \arcsin x$, is defined by $$\arcsin x = y \iff \sin y = x, -\frac{\pi}{2} \leq y \leq \frac{\pi}{2}$$

Since $0$ is the only angle in the interval $[-\frac{\pi}{2}, \frac{\pi}{2}]$ with sine equal to zero, $\arcsin(0) = 0$. Since the sine function has period $2\pi$, knowing that the arcsine of zero is equal to zero tells you that $g$ will have $x$-intercepts at $2\pi$ and $4\pi$. It does not tell you that $g$ also has a zero at $3\pi$. For that value, you can use the observation made above that the sine function is equal to zero when the terminal side of the angle lies on the $x$-axis.

Answer 2

Inverse functions are expressed in radians and degrees. When using a graphics calculator you should choose the proper button... radin/degree selector.

You should be clear that input or argument is ($x$), and result/output is $y$.