MATH 145 – Calculus I
October 26, 2025
Generally, in real life, and in your high school geometry classes, you might have used degrees to measure the size of angles. In calculus, we tend to use the radian instead.
The radian measure of the size of an angle is the length of the arc of a circle of radius \(r = 1\) swept out by the angle, with positive values corresponding to counterclockwise rotation and negative values corresponding to clockwise rotation.
The circumference of the full unit circle is \(C = 2 \pi (1) = 2 \pi\), so \(360^\circ = 2\pi\) radians.
In general, for a circle of radius \(r\), the arc length swept out by an angle of \(\theta\) radians is \(r\theta\).
Using the fact that \(360^\circ = 2\pi\) radians, we can convert from degrees to radians by multiplying the degrees by \(\dfrac{\pi}{180}\), and similarly, we can convert from radians to degrees by multiplying the radians by \(\dfrac{180}{\pi}\).
For an angle \(\theta\), \(\cos(\theta)\) and \(\sin(\theta)\) are the \((x, y)\) coordinates of the intersection of the terminal side of the central angle with radian measure \(\theta\) and initial side on the positive \(x\)–axis.
In other words, the coordinates of the point \(\theta\) units around the unit circle are \(\left( \cos(\theta), \sin(\theta) \right)\).
It follows from this that \(\cos^2(\theta) + \sin^2(\theta) = 1\)
As \(-\infty < t < \infty\), \(\sin(t)\) and \(\cos(t)\) trace out the graph of a function:
These functions are periodic with period \(2\pi\).
\(\sin(t)\) and \(\cos(t)\) (and combinations of them) can be used to model any periodic phenomenon using stretches and shifts of functions, e.g., \(\cos(t) = \sin(t + \frac{\pi}{2})\).
\[ f(x) = A\sin(Bt + C) + D \]
Example: In a port city, high tide is at \(9.9\) feet, and low tide is \(0.1\) feet. High tide is at midnight and low tide is at noon. Find a formula for the height of the tide as a function of hours since midnight.
Other well-known trigonometric functions can be defined in terms of \(\sin(x)\) and \(\cos(x)\):
\[ \begin{aligned} \frac{d}{dx} \sin(x) &= \cos(x) & &\textrm{and} & \frac{d}{dx} \cos(x) &= -\sin(x) \end{aligned} \]
Why? Use the definition of the derivative, the angle sum formula (\(\sin(x + h) = \sin(x)\cos(h) + \cos(x)\sin(h)\)) and some facts about limits: \[ \begin{aligned} \frac{d}{dx} \sin(x) &= \lim\limits_{h \to 0} \frac{\sin(x+h) - \sin(x)}{h} \\ &= \lim\limits_{h \to 0} \frac{\sin(x)\cos(h) + \cos(x)\sin(h) - \sin(x)}{h} \\ &= \sin(x) \left( \lim\limits_{h \to 0} \frac{\cos(h) - 1}{h} \right) + \cos(x) \left( \lim\limits_{h \to 0} \frac{\sin(h)}{h} \right) \\ &= \sin(x) (0) + \cos(x) (1) \\ &= \cos(x) \end{aligned} \]
Examples:
Examples: