Sine and Cosine

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derivatives

In the graph below,

the measurement for the angle $\theta $ is plotted horizontally
$\sin (\theta )$ is the green thicker curve
the blue line is a tangent line to $\sin (\theta )$
the slope of the tangent line is a number and is plotted as the red dot

As you move long the $\sin (\theta )$ curve, the slope of the tangent line follows the red $\cos (\theta )$ curve.

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The values of $\cos (\theta )$ measure the instantaneous rate of change of $\sin (\theta )$.

Derivative of $\sin (\theta )$

$\cos (\theta )$ is the derivative of $\sin (\theta )$.

$\cos (\theta )$ measures the instantaneous rate of change of $\sin (\theta )$.

\[ \frac {d}{d\theta } \sin (\theta ) = \cos (\theta ) \]

In the graph below,

the measurement for the angle $\theta $ is plotted horizontally
$\cos (\theta )$ is the blue curve
the line is a tangent line to $\cos (\theta )$
the slope of the tangent line is a number and is plotted as the red dot

As you move long the $\cos (\theta )$ curve, the slope of the tangent line follows the dotted $-\sin (\theta )$ curve.

This browser does not support embedded elements.

The values of $-\sin (\theta )$ measure the instantaneous rate of change of $\cos (\theta )$.

Derivative of $\cos (\theta )$

$-\sin (\theta )$ is the derivative of $\cos (\theta )$.

$-\sin (\theta )$ measures the instantaneous rate of change of $\cos (\theta )$.

\[ \frac {d}{d\theta } \cos (\theta ) = -\sin (\theta ) \]

We use these two facts as our algebraic reasoning for the behavior of $\sin (\theta )$ and $\cos (\theta )$.

$f(t) = \sin (t)$ is decreasing at $t=\frac {2\pi }{3}$, because $\cos \left ( \frac {2\pi }{3} \right ) < 0$

$g(t) = \cos (t)$ is decreasing at $t=\frac {\pi }{4}$, because $-\sin \left ( \frac {\pi }{4} \right ) < 0$

Shifts

If the graph $y = \sin (t)$ is vertically shifted to $y = \sin (t)+C$, then the shape of the graph does not change. The graph just rigidly moves up. Therefore, the slopes of the tangent lines at the shifted points are the same as for $y = \sin (t)$ .

Vertical Shifts

Shifting a graph vertically doesn’t change the shape. It just repositions the graph. Therefore, all of the slopes of the tangents lines stay the same.

$\blacktriangleright $ The derivative of $\sin (t)+C$ is $\cos (t)$.

$\blacktriangleright $ The derivative of $\cos (t)+C$ is $-\sin (t)$.

\[ \frac {d}{d\theta } (\sin (\theta ) + C) = \cos (\theta ) \]

\[ \frac {d}{d\theta } (\cos (\theta ) + C) = -\sin (\theta ) \]

If the graph $y = \sin (t)$ is horizontally shifted to $y = \sin (t - B)$, then the shape of the graph does not change. The whole graph is just picked up and set down at a new location. Therefore, the slopes of the tangent lines at the shifted points don’t change.

Horizontal Shifts

$\blacktriangleright $ The derivative of $\sin (t - B)$ is $\cos (t - B)$.

$\blacktriangleright $ The derivative of $\cos (t - B)$ is $-\sin (t - B)$.

\[ \frac {d}{d\theta } \sin (\theta - B) = \cos (\theta - B) \]

\[ \frac {d}{d\theta } \cos (\theta - B) = -\sin (\theta - B) \]

Stretching

Suppose you have a line $L$ with slope $m$.

Let $(x_1, y_1)$ and $(x_2, y_2)$ be any two distinct points on $L$.

Then, we have

\[ m = \frac {y_2 - y_1}{x_2 - x_1} \]

Now, suppose we stretch the line vertically by a factor of $k$. That means that all of the points change from $(a, b)$ to $(a, k \cdot b)$

Then, the new slope, $M$, is

\[ M = \frac {k \cdot y_2 - k \cdot y_1}{x_2 - x_1} = \frac {k \cdot (y_2 - y_1)}{x_2 - x_1} = k \cdot \frac {y_2 - y_1}{x_2 - x_1} = k \cdot m \]

If you stretch the graph of a function vertically by a factor of $k$, then you also stretch all of the tangent lines by the same factor.

Vertical Stretching

$\blacktriangleright $ The derivative of $k \cdot \sin (t)$ is $k \cdot \cos (t)$.

$\blacktriangleright $ The derivative of $k \cdot \cos (t)$ is $-k \cdot \sin (t)$.

\[ \frac {d}{d\theta } k \, \sin (\theta ) = k \, \cos (\theta ) \]

\[ \frac {d}{d\theta } k \, \cos (\theta ) = -k \, \sin (\theta ) \]

Let’s put all of this together.

Derivatives

$\blacktriangleright $ The derivative of $A \, \sin (t - B) + C$ is $A \, \cos (t - B)$.

$\blacktriangleright $ The derivative of $A \, \cos (t - B) + C$ is $-A \, \sin (t - B)$.

\[ \frac {d}{d\theta } (A \, \sin (\theta - B) + C) = A \, \cos (\theta - B) \]

\[ \frac {d}{d\theta } (A \, \cos (\theta - B) + C) = -A \, \sin (\theta - B) \]

What is the slope of the tangent line to the graph of $H(t) = -2 \sin \left ( t - \frac {\pi }{6} \right ) + 3$ at $\left ( \frac {5 \pi }{6}, 3-\sqrt {3} \right )$?

The derivative of $f(t) = -2 \sin \left ( t - \frac {\pi }{6} \right ) + 3$ is $f'(t) = -2 \cos \left ( t - \frac {\pi }{6} \right )$.

\[ f'\left ( \frac {5 \pi }{6} \right ) = -2 \cos \left ( \frac {5 \pi }{6} - \frac {\pi }{6} \right ) = -2 \cos \left ( \frac {2 \pi }{3} \right ) = -2 \cdot -\frac {1}{2} = 1\]

The slope of the tangent line is $1$.

Let $G(x) = 3 \cos \left ( x +\frac {\pi }{4} \right ) - 1$.

Where does the graph of $y = G(x)$ have horizontal tangent lines?

We have two approaches:

Maximums and Minimums

The horizontal tangent lines occur at points which correspond to maximums and minimums of cosine.

$\blacktriangleright $ Maximums of cosine functions occur when the inside of the function equals $2 k \pi $ where $k \in \mathbb {Z}$.

\[ x + \frac {\pi }{4} = 2 k \pi \]

\[ x = 2 k \pi - \frac {\pi }{4} = \frac {8 k \pi }{4} - \frac {\pi }{4} = \frac {(8k - 1)\pi }{4} \]

$\blacktriangleright $ Minimums of cosine functions occur when the inside of the function equals $(2 k + 1) \pi $ where $k \in \mathbb {Z}$.

\[ x + \frac {\pi }{4} = (2 k + 1) \pi \]

\[ x = (2 k + 1) \pi - \frac {\pi }{4} = \frac {4(2 k + 1) \pi }{4} - \frac {\pi }{4} = \frac {(8k + 3)\pi }{4} \]

Critical Numbers

The horizontal tangent lines occur at points which correspond to where the derivative equals $0$.

The derivative of $G(x) = 3 \cos \left ( x + \frac {\pi }{4} \right ) - 1$ is $G'(x) = -3 \sin \left ( x + \frac {\pi }{4} \right )$.

The sine function equals $0$ at $k \pi $ where $k \in \mathbb {Z}$.

\[ x + \frac {\pi }{4} = k \pi \]

\[ x = k \pi - \frac {\pi }{4} = \frac {4 k \pi }{4} - \frac {\pi }{4} = \frac {(4k - 1)\pi }{4} \]

Did these two methods give the same set of domain numbers?

\[ \frac {(8k - 1)\pi }{4} : \cdots \frac {-17\pi }{4}, \frac {-9\pi }{4}, \frac {-\pi }{4}, \frac {7\pi }{4} \cdots \]

\[ \frac {(8k + 3)\pi }{4} : \cdots \frac {-13\pi }{4}, \frac {-5\pi }{4}, \frac {3\pi }{4}, \frac {11\pi }{4} \cdots \]

\[ \frac {(4k - 1)\pi }{4} : \cdots \frac {-9\pi }{4}, \frac {-5\pi }{4}, \frac {-\pi }{4}, \frac {3\pi }{4} \cdots \]

Yep. They are generating the same list.

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More Examples of Rates of Change

2025-05-18 00:49:22