Analyzing

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describe everything

Completely analyze $A(\theta ) = \sin (\theta ) - \sin (2\theta )$

$\blacktriangleright $ Domain: $A$ is the difference of two sine functions. Each of those we know to be continuous. Therefore, the implied domain of the sine function is all real numbers.

$\blacktriangleright $ Continuity: The sine function is continuous on the whole real line and $A$ is the difference of two sine functions.

$\blacktriangleright $ Therefore, there are no discontinuities or singularities.

$\sin (\theta )$ is periodic with a period of $2\pi $. $\sin (2\theta )$ is periodic with a period of $\pi $. Therefore, both functions repeat every $2\pi $.

$\blacktriangleright $ Thus, $A$ is periodic with a period of $2\pi $. Therefore, we can just examine one wave. Let’s examine $[0,2\pi )$.

$\blacktriangleright $ Zeros: To locate zeros, we will factor the formula via the double angle sine rule.

\begin{align*} A(\theta ) & = \sin (\theta ) - \sin (2\theta ) \\ A(\theta ) & = \sin (\theta ) - 2\sin (\theta )\cos (\theta ) \\ A(\theta ) & = \sin (\theta ) (1 - 2\cos (\theta )) \\ \end{align*}

Either $\sin (\theta ) = 0$, which happens when $\theta = k\pi $ with $k \in \mathbb {Z}$.
Or, $1 - 2\cos (\theta ) = 0$, which happens when $\theta = \frac {\pi }{3} \pm 2k\pi $ or when $\theta = \frac {5\pi }{3} \pm 2k\pi $ with $k \in \mathbb {Z}$.

In our sample period, we have zeros at $0$ and $\pi $. These will get solid dots as intercepts the graph. There is a (repeat) zero at $2\pi $, but this is not in our sample period. However, we should include a solid dot, because that makes the description clearer. We also have zeros at $\frac {\pi }{3}$and $\frac {5\pi }{3}$. These will get solid dots as intercepts on the graph.

The function is continuous. Therefore, the function either changes sign across each zero or it maintains its sign. The graph either crosses the axis at the intercepts or the graph bounces back and maintains its sign.

We can follow the sign of $A$ through its factors.

\[ A(\theta ) = \sin (\theta ) (1 - 2\cos (\theta )) \]

$\blacktriangleright $ Across $0$, $\sin (\theta )$ changes sign. It goes from negative to positive. On the other hand $1 - 2\cos (\theta )$ stays negative on either side of $0$. Therefore, $A(\theta )$ changes sign. It goes from positive to negative and the graph crosses the axis at $(0, 0)$.

$\blacktriangleright $ Across $\frac {\pi }{3}$, $\sin (\theta )$ remains positive. However, $1 - 2\cos (\theta )$ changes sign. It goes from negative to positive. Therefore, $A(\theta )$ changes sign. It changes from negative to positive and the graph crosses the axis at $\left ( \frac {\pi }{3}, 0 \right )$.

$\blacktriangleright $ Across $\pi $, $\sin (\theta )$ changes sign, from positive to negative. $1 - 2\cos (\theta )$ stays positive on either side of $\pi $. Therefore, $A(\theta )$ changes sign from positive to negative and the graph crosses the axis at $(\pi , 0)$.

$\blacktriangleright $ Across $\frac {5\pi }{3}$, $\sin (\theta )$ remains negative. However, $1 - 2\cos (\theta )$ changes sign. It goes from positive to negative. Therefore, $A(\theta )$ changes sign. It changes from negative to positive and the graph crosses the axis at $\left ( \frac {5\pi }{3}, 0 \right )$.

The behavior at $2\pi $ is the same as the behavior at $0$, since $A$ is periodic.

We now know that $A(t)$ is negative when $t$ is slightly positive and that $A$ changes signs at each zero. We can sketch a rough idea of how the graph moves.

$\blacktriangleright $ The range is difficult to determine from here. It depends on how high and low the maximum and minimum are.

We do know that $\sin (\theta )$ and $\cos (\theta )$ can’t get greater than $1$ or less than $-1$. That tells us that $1 -2\cos (\theta )$ can’t get greater than $3$. So, the range must be inside $[-3, 3]$.

A better graph seems to agree with this reasoning.

The graph suggests a lot to us (which is why we like graphs). The graph gives us reason to believe that the range is inside the interval $[-2, 2]$. However, we’ll need Calculus to get exact values for the maximum and minimum values.

The graph suggests a lot to us (which is why we like graphs). The graph also suggests that there are critical numbers in between the zeros.

$\blacktriangleright $ maximums and minimums

$A$ has a critical number between $\frac {2\pi }{3}$ and $\pi $, which corresponds to a global (and local) maximum.
$A$ has a critical number between $\pi $ and $\frac {5\pi }{3}$, which corresponds to a global (and local) minimum.
$A$ has a critical number between $\frac {5\pi }{3}$ and $2\pi $, which corresponds to a local maximum.
$A$ has a critical number between $0$ and $\frac {2\pi }{3}$, which corresponds to a local minimum.

Let’s call these critical numbers $c_1$, $c_2$, $c_3$, and $c_4$.

$\blacktriangleright $ Rate-of-Change

$A$ is decreasing on $[0, c_1]$.
$A$ is increasing on $[c_1, c_2]$.
$A$ is decreasing on $[c_2, c_3]$.
$A$ is increasing on $[c_3, c_4]$.
$A$ is decreasing on $[c_4, 2\pi ]$.

Of course, we could use the graph to approximate values for these critical numbers.

with Calculus

Calculus will give us the tools to discover a formula for the derivative of $A$.

\[ A'(\theta ) = \cos (\theta )-2 \cos (2\theta ) \]

The critical numbers would be where $A'(\theta ) = 0$.

\begin{align*} 0 & = \cos (\theta )-2 \cos (2\theta ) \\ & = \cos (\theta ) - 2 (2 \cos ^2(\theta ) - 1) \\ & = -4 \cos ^2(\theta ) + \cos (\theta ) + 2 \\ & = 4 \cos ^2(\theta ) - \cos (\theta ) - 2 \\ \end{align*}

We have a quadratic equation. It doesn’t factor nicely, so we’ll go with the quadratic formula.

\[ \cos (\theta ) = \frac {1 \pm \sqrt {1 - 4 (4)(-2)}}{2 \cdot 4} = \frac {1 \pm \sqrt {33}}{8} \]

Let’s take each of these separately.

$\blacktriangleright $ $\cos (\theta ) = \frac {1 + \sqrt {33}}{8} \approx 0.843$

We are looking for an angle whose cosine is $\frac {1 + \sqrt {33}}{8}$. This is positive, therefore, there are two such angles. One in the first quadrant. That would be $\arccos \left (\frac {1 + \sqrt {33}}{8}\right )$

There is a second angle in the fourth quadrant where cosine is $\frac {1 + \sqrt {33}}{8}$. This angle could be described as $-\arccos \left (\frac {1 + \sqrt {33}}{8}\right )$. However, our chosen interval to examine is $[0, 2\pi )$. So, the angle in the fourth quadrant is $2\pi - \arccos \left (\frac {1 + \sqrt {33}}{8}\right )$

$\blacktriangleright $ $\cos (\theta ) = \frac {1 - \sqrt {33}}{8} \approx -0.593$

We are looking for an angle whose cosine is $\frac {1 - \sqrt {33}}{8}$. This is negative, therefore, there are two such angles. One in the second quadrant. That would be $\arccos \left (\frac {1 - \sqrt {33}}{8}\right )$

There is a fourth angle in the third quadrant where cosine is $\frac {1 - \sqrt {33}}{8}$. This third quadrant angle has the same reference angle as $c_2$. Therefore, $c_3 = 2\pi - \arccos \left (\frac {1 - \sqrt {33}}{8}\right )$.

We want exact answers, when possible, and we have exact answers with Calculus.

For us, without the derivative, we could approximate the critical numbers from a graph as well as the extreme values of $A$.

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$c_1 \approx 0.568$
$c_2 \approx 2.206$
$c_3 \approx 4.078$
$c_4 \approx 5.715$

$\blacktriangleright $ Extrema

$\rhd $ We have a global maximum of $A(c_2)$:

\[ A(c_2) = \sin \left (\arccos \left (\frac {1 - \sqrt {33}}{8}\right )\right ) - \sin \left (2 \arccos \left (\frac {1 - \sqrt {33}}{8}\right )\right ) \approx 1.76 \]

at $c_2 = \arccos \left (\frac {1 - \sqrt {33}}{8}\right ) \approx 2.206$.

$\rhd $ We have a global minimum of $A(c_3)$:

\[ A(c_3) = \sin \left ( 2\pi - \arccos \left (\frac {1 - \sqrt {33}}{8}\right ) \right ) - \sin \left (2 \left ( 2\pi - \arccos \left (\frac {1 - \sqrt {33}}{8}\right ) \right ) \right ) \approx -1.76 \]

at $c_3 = 2\pi - \arccos \left (\frac {1 - \sqrt {33}}{8}\right ) \approx 4.078$.

$\rhd $ We have a local minimum of $A(c_1)$:

\[ A(c_1) = \sin \left ( \arccos \left (\frac {1 + \sqrt {33}}{8}\right ) \right ) - \sin \left (2 \arccos \left (\frac {1 + \sqrt {33}}{8}\right ) \right ) \approx -0.369 \]

at $c_1 = \arccos \left (\frac {1 + \sqrt {33}}{8}\right ) \approx 0.568$.

$\rhd $ We have a local maximum of $A(c_4)$:

\[ A(c_4) = \sin \left ( 2\pi - \arccos \left (\frac {1 + \sqrt {33}}{8}\right ) \right ) - \sin \left (2 (2\pi - \arccos \left (\frac {1 + \sqrt {33}}{8}\right )) \right ) \approx 0.369 \]

at $c_4 = 2\pi - \arccos \left (\frac {1 + \sqrt {33}}{8}\right ) \approx 5.715$.

$\blacktriangleright $ Rate-of-Change

$A$ is decreasing on approximately $[0, 0.568]$.
$A$ is increasing on approximately $[0.568, 2.206]$.
$A$ is decreasing on approximately $[2.206, 4.078]$.
$A$ is increasing on approximately $[4.078, 5.715]$.
$A$ is decreasing on approximately $[5.715, 2\pi ]$.

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2025-05-18 00:10:17