Analysis

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sine

Analyze $T(r) = 2 \sin (3r - \pi ) - 1$

We are going to view this as a composition of three functions.

\[ T = Out \circ \sin \circ In \]

where

$Out(x) = 2x - 1$

$In(y) = 3y - \pi $

Along with $\sin (\theta )$, these three component functions, we get

\[ T(r) = (Out \circ \sin \circ In)(r) = Out(sin(In(r))) = 2 \sin (3r - \pi ) - 1 \]

$T$ is a periodic function, since $\sin (\theta )$ is a periodic function.

The principal interval of $\sin (\theta )$ is $[0, 2\pi )$. The principal interval of $T$ will be the values of $r$ that make the values of $In(y)$ run from $0$ to $2\pi $.

$3y - \pi = 0$ at $y = \frac {\pi }{3}$

$3y - \pi = 2\pi $ at $y = \pi $

The principle interval of $T$ is $\left [ \frac {\pi }{3}, \pi \right )$.

The length of this interval is $\frac {2\pi }{3}$, which is the period of $T$.

$\blacktriangleright $ desmos graph

This browser does not support embedded elements.

$T$ is a periodic function with period $\frac {2\pi }{3}$.

Therefore, our analysis will focus on the principal interval of $\left [ \frac {\pi }{3}, \pi \right )$.

All of the features and characteristics we discover will repeat wiht a period of $\frac {2\pi }{3}$.

Domain

The domain of $Out(x) = 2x - 1$ is $(-\infty , \infty )$, because $Out$ is a linear function.

This includes any value of $\sin (\theta )$. That means we can use the entire domain of $\sin (\theta )$, which is $(-\infty , \infty )$.

This includes any value of $In(y)$, which means we can use the entire domain of $In(y)$, which is $(-\infty , \infty )$, because $In$ is a linear function.

The domain of $T$ is $(-\infty , \infty )$.

Zeros

Since the values of $T$ come from the values of $Out$, we are first looking for zeros of $Out$. $Out(x) = 2x - 1$ is a linear function and has only one zero, $\frac {1}{2}$.

We are looking for where $\sin (\theta ) = \frac {1}{2}$.

There are two numbers in $[0, 2\pi )$ where $\sin (\theta ) = \frac {1}{2}$. They are $\frac {\pi }{6}$ and $\frac {5\pi }{6}$.

We need the values of $y$ where $In(y) = 3y - \pi = \frac {\pi }{6}$ and $In(y) = 3y - \pi = \frac {5\pi }{6}$

$y = \frac {7\pi }{18}$ and $y = \frac {11\pi }{18}$

[ These agree with the graph. ]

And, these repeat every $\frac {2\pi }{3}$ for $T$.

Continuity

The component functions of the composition are linear and sine, all continuous.

The composition of continuous function is continuous.

$T$ is continuous. It has no disontinuities.

Since the domain is $(-\infty , \infty )$, there are no singularities.

End-Behavior

$T$ is a periodic function with period $\frac {2\pi }{3}$.

It either has no end-behavior or the end-behavior is that it is periodic.

Behavior

We need the behavior of each of the three component funcitons and then we will compose them together.

$Out(x) = 2x - 1$ is a linear function with a positive leading coefficient, so it is an increasing function.

$In(y) = 3y - \pi $ is a linear function with a positive leading coefficient, so it is an increasing function.

$\sin (\theta )$ increases and decreases through the quadrants. For the principal interval, we have

increases on $\left ( 0, \frac {\pi }{2} \right )$
decreases on $\left ( \frac {\pi }{2}, \pi \right )$
decreases on $\left ( \pi , \frac {3\pi }{2} \right )$
increases on $\left ( \frac {3\pi }{2}, 2\pi \right )$

Now to trace domains and ranges.

We need the values of $In(y) = 3y - \pi $ to be $0$, $\frac {\pi }{2}$, $\pi $, $\frac {3\pi }{2}$, and $2\pi $.

$\blacktriangleright $ $3y - \pi = 0$ when $y = \frac {\pi }{3}$

$\blacktriangleright $ $3y - \pi = \frac {\pi }{2}$ when $y = \frac {\pi }{2}$

$\blacktriangleright $ $3y - \pi = \pi $ when $y = \frac {2\pi }{3}$

$\blacktriangleright $ $3y - \pi = \frac {3\pi }{2}$ when $y = \frac {5\pi }{6}$

$\blacktriangleright $ $3y - \pi = 2\pi $ when $y = \pi $

Now to glue everything together.

On $r = y \in \left ( \frac {\pi }{3}, \frac {\pi }{2} \right )$,

$In(y) = 3y - \pi $ is increasing. The range of $In(y)$ is $\theta = In(y) \in \left (0, \frac {\pi }{2} \right )$, where $\sin (\theta )$ is increasing.

The range of $\sin (\theta )$ on $\theta \in \left (0, \frac {\pi }{2} \right )$ is $x = \sin (\theta ) \in (0, 1)$, where $Out(x)$ is increasing, because $Out$ is always increasing.

\[ T(r) = increasing \circ increasing \circ increasing = increasing \]

On $r = y \in \left ( \frac {\pi }{2}, \frac {2\pi }{3} \right )$,

$In(y) = 3y - \pi $ is increasing. The range of $In(y)$ is $\theta = In(y) \in \left (\frac {\pi }{2}, \pi \right )$, where $\sin (\theta )$ is decreasing.

The range of $\sin (\theta )$ on $\theta \in \left ( \frac {\pi }{2}, \pi \right )$ is $x = \sin (\theta ) \in (0, 1)$, where $Out(x)$ is increasing, because $Out$ is always increasing.

\[ T(r) = increasing \circ decreasing \circ increasing = decreasing \]

On $r = y \in \left ( \frac {2\pi }{3}, \frac {5\pi }{6} \right )$,

$In(y) = 3y - \pi $ is increasing. The range of $In(y)$ is $\theta = In(y) \in \left (\pi , \frac {3\pi }{2} \right )$, where $\sin (\theta )$ is decreasing.

The range of $\sin (\theta )$ on $\theta \in \left ( \pi , \frac {3\pi }{2} \right )$ is $x = \sin (\theta ) \in (-1, 0)$, where $Out(x)$ is increasing, because $Out$ is always increasing.

\[ T(r) = increasing \circ decreasing \circ increasing = decreasing \]

On $r = y \in \left ( \frac {5\pi }{6}, \pi \right )$,

$In(y) = 3y - \pi $ is increasing. The range of $In(y)$ is $\theta = In(y) \in \left ( \frac {3\pi }{2}, 2\pi \right )$, where $\sin (\theta )$ is increasing.

The range of $\sin (\theta )$ on $\theta \in \left ( \frac {3\pi }{2}, 2\pi \right )$ is $x = \sin (\theta ) \in (-1, 0)$, where $Out(x)$ is increasing, because $Out$ is always increasing.

\[ T(r) = increasing \circ increasing \circ increasing = increasing \]

This behavior repeats with a period of $\frac {2\pi }{3}$.

[ This agrees with the graph. ]

$\blacktriangleright $ desmos graph

This browser does not support embedded elements.

Local Maximum and Minimum

$T$ is continuous on $(-\infty , \infty )$.

$\blacktriangleright $ $T$ switches from increasing to decreasing at $\frac {\pi }{2}$, which makes $\frac {\pi }{2}$ a crtical number and the location of a local maximum.

$\blacktriangleright $ $T$ switches from decreasing to increasing at $\frac {5\pi }{6}$, which makes $\frac {5\pi }{6}$ a crtical number and the location of a local minimum.

These repeat with a period of $\frac {2\pi }{3}$.

\[ T\left ( \frac {\pi }{2} \right ) = 2 \sin \left (3 \cdot \frac {\pi }{2} \right ) - 1 = -3 \]

\[ T\left ( \frac {5\pi }{6} \right ) = 2 \sin \left (3 \cdot \frac {5\pi }{6} \right ) - 1 = 1 \]

Global Maximum and Minimum

Since $T$ is continuous on $(-\infty , \infty )$, $\frac {\pi }{2}$ and $\frac {5\pi }{6}$ are the only critical numbers, we also have global extrema here.

\[ T\left ( \frac {\pi }{2} \right ) = 2 \sin \left (3 \cdot \frac {\pi }{2} \right ) - 1 = -3 \]

\[ T\left ( \frac {5\pi }{6} \right ) = 2 \sin \left (3 \cdot \frac {5\pi }{6} \right ) - 1 = 1 \]

These repeat with a period of $\frac {2\pi }{3}$.

Range

Since $T$ is continuous with a global maximum of $1$ and a global minimum of $-1$, the range is $[-3, 1]$.

[ This agrees with the graph. ]

Periodic

$T$ is periodic with period $\frac {2\pi }{3}$.

We have zeros in the principal interval as $\frac {7\pi }{18}$ and $\frac {11\pi }{18}$.

To describe all of the zeros of $T$, we need to include all numbers which are these two numbers plus or minus and whole number of $\frac {2\pi }{3}$.

\[ \left \{ \frac {7\pi }{18} + k \cdot \frac {2\pi }{3} \, | \, k \in \mathbb {Z} \right \} \]

\[ \left \{ \frac {11\pi }{18} + k \cdot \frac {2\pi }{3} \, | \, k \in \mathbb {Z} \right \} \]

Similarly, we could describe the intervals by including $k \cdot \frac {2\pi }{3} $.

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2025-08-08 13:36:47