Trigonometry: A day at the track
Inverse Trigonometric Functions
UBC Calculus Online Course Notes

Derivatives of Trigonometric functions

Let's begin by making a few observations about the functions $ \sin(t) $ and $ \cos(t) $ . From now on, you will hopefully think of these functions as the y and x coordinates of a point moving around and around a circle.

For your consideration:

    • What is the radius of the circle if its x and y coordinates are given by $  x=cos(t)  $ and $  y=sin(t)  $ ?
    • How many revolutions are made around the circle per unit time? (Note: a revolution occurs every multiple of $  2 \pi $ )

    • Which of the two functions has a "flat" portion of its graph at $  t=0  $ (remember that flat means slope zero.)
    • Where does each of the functions have maximum values? What are those maximum values? (For example, $  \cos(t)  $ has a maximum value of 1 at t=0 and where else?).
    • Carefully observe the graph of $  \sin(t)  $ . How does the slope of the curve change as t increases? Sketch a plot which would show how the slope of the curve changes as t increases. (Use only the information in the graphs.)

As have done previously, we can try to sketch the graphs of the derivatives of these two functions just using information about where the functions are increasing and decreasing. If we do that, we find pictures like this:

What do you notice about these graphs? Hopefully, the remarkable fact that the derivative of the sine function resembles the cosine function. Also the derivative of the cosine function seems related to the sine function. We will use a calculation to verify this relationship below. Later, we will see that some interesting phenomena arise because of the fact that the derivative of a trigonometric function is another trigonometric function.


The derivative of $  \sin(t)  $

We will begin by determining the derivative at just one point. Notice that

 \begin{eqnarray*} 
\frac{d}{dt} \sin(t) |_{t=0} & = & \lim_{h \to 0} \frac{\sin(h) - 
\sin(0)}{h} \\ 
& = & \lim_{h\to 0} \frac{\sin(h)}{h} 
\end{eqnarray*}

Now the problem is that this is a difficult limit to evaluate. The sine function is complicated to compute and here we are asking for a delicate limit: it does not work to simply substitute $  h = 0 
 $ into the expression because it would then be undefined.

We can, however, try to understand this limit another way. This limit represents the derivative of the sine function at $  t = 
0 $ . We could try to zoom in on the graph of the sine function and find out what its slope is at that point.

This shows us that the derivative of the sine function is 1 at $  t = 0  $ . In other words,

\[  \frac{d}{dt} \sin(t)|_{t=0} = \lim_{h\to 0}\frac{\sin(h)}{h} = 1 
 \]

While we're at it, we can also compute the same quantity for the cosine function.

\[ 
\frac{d}{dt}\cos(t)|_{t=0} = \lim_{h\to 0}\frac{\cos(h) - \cos(0)}{h} = 
\lim_{h\to 0} \frac{\cos(h) - 1}{h} 
 \]

To determine this derivative, we can again zoom in on the graph:

In other words,

\[ 
\frac{d}{dt}\cos(t)|_{t=0} = \lim_{h\to 0}\frac{\cos(h) - 1}{h} = 0 
 \]

Now we are in good position to compute the derivative of the sine function. The crucial ingredient is the angle addition formula.

 \begin{eqnarray*} 
\frac{d \sin(t)}{dt} & = & \lim_{h\to 0}\frac{\sin(t+h) - \sin(t)}{h} \\ 
& = & \lim_{h\to 0} \frac{\sin(t)\cos(h) + \cos(t)\sin(h) - \sin(t)}{h} \\ 
& = & \lim_{h\to 0} \frac{\sin(t) (\cos(h) - 1) + \cos(t) \sin(h)}{h} \\ 
& = & \sin(t) \lim_{h\to 0}\frac{\cos(h) - 1}{h} + \cos(t) \lim_{h\to 
0} \frac{\sin(h)}{h} \\ 
& = & \cos(t) 
\end{eqnarray*}

Here we have verified our observation that the derivative of the sine function is related to the cosine function. In fact, we have found that it is exactly equal to the cosine function.

To determine the derivative of the cosine function, we can remember that

 \begin{eqnarray*} 
\cos(t) & = & \sin(\frac \pi 2 - t) \\ 
\sin(t) & = & \cos(\frac \pi 2 - t) 
\end{eqnarray*}

Then it follows that


\begin{eqnarray*} 
\frac{d}{dt}\cos(t) & = & \frac{d}{dt} \sin(\frac \pi 2 - t) \\ 
& = & -\cos(\frac \pi 2 - t) \\ 
& = & -\sin(t) 
\end{eqnarray*}

This again verifies the intuition we gained from roughly sketching the derivative above. We conclude that

 \begin{eqnarray*} 
\frac{d}{dt} \sin(t) & = & \cos(t) \\ 
\frac{d}{dt} \cos(t) & = & -\sin(t) 
\end{eqnarray*}


Other derivatives

From these two derivatives, we can compute the derivatives for the other trigonometric functions using our now standard tools.

 \begin{eqnarray*} 
\frac{d}{dt} \tan(t) = \frac{d}{dt} \frac{\sin(t)}{\cos(t)} = 
\sec^2(t) & & 
\frac{d}{dt} \sec(t) = \sec(t)\tan(t) \\ 
\frac{d}{dt} \cot(t) = -\csc^2(t) & & \frac{d}{dt} \csc(t) = 
-\csc(t)\cot(t) 
\end{eqnarray*}


The importance of radians

It is important to note that our use of radians here is crucial. If we were to measure the argument of the sine function in, say, degrees, we would find a different result. The following graphs make this clear. On the left is the graph of the sine function in radians while on the right it is graphed in degrees. Notice that the rate of change of the two graphs is much different. In calculus, we use radians because they give such a nice result for the derivative of the sine and cosine function.