Derviatives of the Trigonmetric Functions

Let's consider differentiating the sine function:

Before we continue, recall these two propositions we saw in the previous section:

1. ${\lim\limits_{x \to 0} \dfrac{\sin x}{x} = 1}$
2. ${\lim\limits_{x \to 0} \dfrac{1 - \cos x}{x} = 0}$

Keep these propositions in mind as we derive the derivatives of the trigonometric functions. Now, to differentiate ${\sin x,}$ we use the familiar difference quotient, and apply the limit:

As we saw in the section on limits, we want to reframe this expression. To do so, we use a trigonometric identity:

Applying this identity:

We cannot just plug in 0 into this limit — it's one of the trickier ones. We must try and group the terms. To group terms, the first thing we want to do is gather what we know. From basic trigonometry, we know that ${\lim\limits_{\Delta x \to 0} \cos(\Delta x)}$ is 1. We also see that ${\sin x}$ is a common term. Using algebra:

Now, if we look at the final result from a broad overview, this is really just:

where

• ${m = \left( \dfrac{\cos (\Delta x) - 1}{\Delta x} \right),}$ and
• ${m = \left( \dfrac{\sin \Delta x}{\Delta x} \right)}$

We need a way to get rid of ${m}$ and ${n.}$ To do so, we will use the following properties, to be proven later:

1. ${\lim\limits_{\Delta x \to 0} \dfrac{\cos \Delta x - 1}{\Delta x} = 0}$
2. ${\lim\limits_{\Delta x \to 0} \dfrac{\sin \Delta x}{x} = 1}$

Using these properties, we have:

Above, we have ${(\sin x \cdot 0) + (\cos x \cdot 1).}$ This gives us ${\cos x.}$ Thus, we now have a specific formula for the derivative of ${sin x:}$

Let's now consider the derivative of ${\cos x.}$ The process is similar. Again, we start with the diffference quotient:

Just as we used a trigonometric identity in deriving the derivative of ${\sin x,}$ we will use another trionometric identity for ${\cos x,}$ the angle sum formula:

Applying the angle sum formula, we have:

Again, we see that this is really just:

where,

• ${a = \dfrac{\cos \Delta x - 1}{\Delta x}}$ and
• ${b = \dfrac{ \sin \Delta x}{\Delta x}}$

We want to get rid of ${a}$ and ${b,}$ so we use the same properties we saw earlier:

1. ${\lim\limits_{\Delta x \to 0} \dfrac{\cos \Delta x - 1}{\Delta x} = 0}$
2. ${\lim\limits_{\Delta x \to 0} \dfrac{\sin \Delta x}{x} = 1}$

Applying the two properties:

Thus, we know have a specific formula for the derivative of ${\cos x:}$

As an aside, let's think about these formulas a little more carefully to contextualize these derivations. Consider the derivative of ${\cos x.}$ When we evaluate ${\dfrac{d}{dx} \cos x}$ at ${x = 0,}$ then by the definition of a derivative, this is ${\lim\limits_{\Delta x \to 0} \dfrac{\cos (\Delta x) - 1}{\Delta x}.}$ This limit evaluates to 0, as we saw earlier.

The same phenomenon appears for the derivative of ${\sin x.}$ When we evaluate ${\dfrac{d}{dx} \sin x,}$ we are just evaluating ${\lim\limits_{\Delta x \to 0} \dfrac{\sin \Delta x}{\Delta x}.}$ This limit evaluates to 1, again, as we saw earlier.

What this tells us is that the derivatives of ${\sin x}$ and ${\cos x}$ at ${x = 0}$ yields all the values of ${\dfrac{d}{dx} \sin x}$ and ${\dfrac{d}{dx} \cos x.}$

To actually understand how we've derived the specific formulas above, we must understand the behavior of ${\dfrac{\sin x}{x}}$ and ${\dfrac{\cos x - 1}{x}}$ when ${x}$ is close to 0. In other words, we must prove the properties we've been using:

1. ${\lim\limits_{\Delta x \to 0} \dfrac{\cos \Delta x - 1}{\Delta x} = 0}$

2. ${\lim\limits_{\Delta x \to 0} \dfrac{\sin \Delta x}{x} = 1}$

To construct these proofs, we must use geometry, because ${\sin x}$ and ${\cos x}$ are geometric propositions. To do so, we'll need to use a variable other than ${\Delta x.}$ Namely, ${\theta.}$ We do so because we want to go back to thinking about sine and cosine as trigonometric ratios, rather than functions. We'll first consider ${\dfrac{\sin x}{x}.}$ Accordingly, what we are really trying to prove is:

Let's first start by looking at the unit circle:

Here, our proof will depend on the followng proposition: When the radius of the circle has as length of 1, ${\theta}$ is the length of the arc above (highlighted red). Note that this proposition is true only if the angle ${\theta}$ is measured in radians. It is not true if ${\theta}$ is measured in degrees. This an example of why prefer measuring angles in radians rather than degrees in mathematics. Recall the definition of sine:

Since the radius is of length 1, then the hypotenuse is of length 1. Thus, the length of the side opposite ${\theta}$ has a length of ${\sin \theta.}$ Now, for the sake of argument, let's make a copy of that triangle and reflect it:

From the circle above, we see that the total arc length is now ${2 \theta.}$ We also see that we've formed an isosceles triangle, where the green side is of length ${2 \sin \theta,}$ while the blue sides are of length ${1.}$ We also see that the interior angle has a measure of ${2 \theta.}$ With these lengths, the ratio of the green edge of the triangle to the arc length is:

Now the question is, why does ${\dfrac{\sin \theta}{\theta}}$ tend to 1 as ${\theta}$ gets closer and closer to 0? Because as the angle ${\theta}$ gets very very small, the arc looks more and more like a straight line. Moreover, as ${\theta}$ gets closer and closer to 0, the green segment and the red arch start to merge.

This shows that short curves are nearly straight. Hence, we have:

Now let's consider the property ${\lim\limits_{x \to 0} \dfrac{\cos x - 1}{x} = 0.}$ Again, we want to use ${\theta.}$ Here, however, because we are thinking about sine and cosine geometrically, we want a positive quantity, since we're dealing with lengths. So, we perform some manipulation up front:

Thus, the hypothesis we want to prove is:

The -1 is just a constant, so let's just clean it up a bit more by multiplying ${\dfrac{1}{-1}}$ to both sides:

To prove the hypothesis above, we focus on the “gap” between the edge and the arc. Since the distance between the origin and the arc is 1, this gap has a length of ${1 - \cos \theta.}$

Just as we saw with sine, as the angle ${\theta}$ gets very small, the distance ${1 - \cos \theta}$ gets very, very small.

At some point, the segment ${1 - \cos \theta}$ is 0. Hence, we have the following:

Note that the limit isn't simply ${\dfrac{0}{0}.}$ This is an example of a limit where we aren't just doing a simple plug-and-play. The ${\theta}$ in the denominator is the arc length. The ${\theta}$ we're taking to 0 is the interior angle. Thus, making the angle ${\theta}$ smaller and smaller, the arc length ${\theta}$ is getting closer and closer to 0, but not nearly as fast as ${1 - \cos \theta.}$ Remember, the property we're trying to prove is a ratio. Thus, because ${\cos \theta - 1}$ approaches 0 faster than the arc length ${\theta,}$ the limit of the ratio is 0. To think about this more clearly, compare the approximations in the table above. Notice how much faster ${1 - \cos \theta}$ approaches 0 compared to ${\theta.}$

The proof above is fairly abstract. Let's see another proof. First, suppose there is a point ${P}$ on a unit circle, then move along to ${Q.}$

From the circle above, we see that ${\sin \theta}$ is the vertical distance between ${P}$ and the {\text{x-axis}.} The ${\Delta \theta}$ is the change in the angle, as we move from ${P}$ to ${Q.}$ Accordingly, ${Q}$ is the point on the unit circle where the angle measures ${\theta + \Delta \theta.}$ It follows then that the {\text{y-coordinate}} of ${Q}$ is ${\sin (\theta + \Delta \theta).}$ To determine the rate of change of ${\sin \theta}$ with respect to theta (i.e., ${\dfrac{d}{d \theta} \sin \theta,}$) we must determine the rate of change for ${y = \sin \theta.}$

If we zoomed in on the ${\Delta y}$ above:

The curved portion is the part of the circle. We've also drawn a line segment from ${P}$ to ${Q.}$ In doing so, we've drawn a right triangle ${\triangle PQR.}$ Now, we use a concept we saw earlier: Very short pieces of curves are nearly straight. This means that the ${\text{arc } PQ}$ is essentially the same as the edge ${\overline{\rm PQ}:}$

Now, we know that ${\text{arc } PQ}$ has a length of ${\Delta \theta.}$ Thus, we know that:

Great, we have one side down. If we can determine the measure of ${\angle QPR}$ is, then we can determine ${\Delta y.}$ Geometrically, we are trying to find the length of ${\overline{\rm PR.}}$ To think about what the measure of ${\angle QPR}$ might be, let's see the line with the whole unit circle:

Notice that the segment is almost perpendicular to the radius ${\overline{\rm OP}.}$ When we take the limit, it is just about perpendicular:

If we rotate the entire construction by ${{90}^\circ,}$ we can see that ${\angle QPR}$ is roughly the angle ${\theta.}$

Thus, we can conclude:

Accordingly:

With these two propositions: (1) ${\overline{\rm PQ} \approx \Delta \theta,}$ and (2) ${\angle QPR \approx \theta,}$ we can now draw the following premise:

It is this premise that allows us to reach the proposition we were after:

And by applying the limit: