Monthly Archives: October 2018

The Derivative of the Sine and Cosine Functions

In AP Calculus, we saw the following derivatives: \[\sin’ x = \cos x \\ \cos’ x = -\sin x\]

That is, the derivative of the sine is the cosine, and the derivative of the cosine is the opposite of the sine.

I showed one way to see why this is true using Desmos. The purple line is the tangent, while the black dot is the slope of that tangent. As you move the slider a from side to side, the black dot traces the cosine function.

The book has an algebraic proof of the first rule (p. 112), using the limit process. In this post, I’ll discuss a more visual, geometry-based explanation.

We’ll start with a few triangles and a portion of the unit circle. 

The angle \(\theta\) represents our original angle, while \(\delta = \Delta\theta\). Note that we are NOT looking at the change in the \(x\) value; we’re looking at the change in the angle measurement. This is key!

So what we want to know is: How does \(\sin\theta\) change as \(\theta\) changes?

\(\overline{AC}\) is the radius of the unit circle, so its length is 1; \(\sin\theta = m\overline{CF}/m\overline{AC} = f\). Likewise, \(\cos\theta = g\).

\(\overline{EC}\) is on the tangent to the circle at point \(C\). \(\overline{CD}\) is collinear with \(\overline{CF}\); \(\overline{ED}\perp\overline{CD}\).

So let’s consider the blue and green triangles. Using geometry, we know that since the blue triangle is a right triangle, \(\theta\) and \(\gamma\) are complementary. Since \(\overline{DF}\) is on a line, and since \(\angle ECA\) is right, \(\alpha\) and \(\gamma\) are complementary… meaning that \(\alpha\cong\theta\).

More about the green triangle: \(\cos\alpha = \cos\theta = e/c \Rightarrow e = c \cos\theta\). But \(c\) is also opposite \(\delta\) in \(\triangle ECA\), meaning that \(\sin\delta = c/a \Rightarrow c = a\sin\delta\) and \(e = a\sin\delta\cos\theta\).

We’re ready to write our limit now. We want to know what happens to \(e\) when \(\delta\) gets smaller and smaller. Specifically: \[\lim_{\Delta\theta\rightarrow 0}\frac{\Delta y}{\Delta\theta} = \lim_{\delta\rightarrow 0}\frac{e}{\delta} \\ = \lim_{\delta\rightarrow 0}\frac{a\sin\delta\cos\theta}{\delta} \\ = \lim_{\delta\rightarrow 0} a \cdot \lim_{\delta\rightarrow 0}\frac{\sin\delta}{\delta} \cdot \lim_{\delta\rightarrow 0} \cos\theta\]

Consider each in turn. As \(\delta\) gets smaller, the difference between \(a\) and the radius of the unit circle gets smaller. That is, \[\lim_{\delta\rightarrow 0} a = 1\]

From our examination of the Squeeze Theorem, we know the important identity \[\lim_{x\rightarrow 0}\frac{\sin x}{x} = 1\] which we can apply here (where \(x = \delta\)).

Finally, since \(\cos\theta\) is not affected by \(\delta\) at all, \[\lim_{\delta\rightarrow 0} \cos\theta = \cos\theta\]

Putting these together, we get  \[\lim_{\delta\rightarrow 0} a \cdot \lim_{\delta\rightarrow 0}\frac{\sin\delta}{\delta} \cdot \lim_{\delta\rightarrow 0} \cos\theta = 1 \cdot 1 \cdot \cos\theta = \cos\theta\] QED.

The process for demonstrating the cosine is very similar (using \(d = a\sin\delta\sin\theta\)), but the question that comes up is: Why is the cosine’s derivative the opposite of the sine? Notice what happens in the green triangle: While the height goes upward, the width (corresponding to \(\sin\alpha = \sin\theta\)) goes to the left. This is why the cosine’s derivative is the opposite of the sine.