# MSC Notes: Computing the Derivative of Cosine In Various Forms

On Wednesday, when I was tutoring at the Math Study Center at Phillips Academy, I came across the following limit: \[\lim_{x \to a} \frac{\cos x - \cos a}{x - a}.\] We know that the limit is the actually the derivative of cosine at the point \(x = a\), which should evaluate to \(-\sin a\). Furthermore, we know how to evaluate more popular form of the derivative, \[\lim_{h \to 0} \frac{\cos(a + h) - \cos a}{h},\] obtained by making the substitution \(h = x - a\). The evaluation of the latter limit amounts to applying the cosine angle addition identity and then factoring out a few special trigonometric limits, a standard textbook solution. In the spirit of the problem, I tried to compute the former (given) limit without the substitution, and I discovered a quite simple and elegant solution that involves using a trigonometric identity that can be derived using the cosine angle sum and difference identities.

### The Solution

We will be solving the definition limit of the derivative of cosine in the following form \[\lim_{x \to a} \frac{\cos x - \cos a}{x - a}\] without the use of substitution that changes it to the more popular form \[\lim_{h \to 0} \frac{\cos(a + h) - \cos a}{h}.\]

We will need a special sum-to-product trigonometric identity to solve this limit. To review, \[\begin{cases} \cos(p - q) = \cos p \cos q + \sin p \sin q, \\ \cos(p + q) = \cos p \cos q - \sin p \sin q.\end{cases}\] Notice that subtracting the second equation from the first yields \[\cos(p-q) - \cos(p+q) = 2 \sin p \sin q. \]

This identity equates the difference of cosines to a product of sines, which seems promising for our limit. If we set \(p = \tfrac{x+a}{2}\) and \(q = \tfrac{x-a}{2}\). Then, \[p - q = \frac{x+a}{2} - \frac{x-a}{2} = a, \, \mathrm{and} \; p + q = \frac{x+a}{2} + \frac{x-a}{2} = x.\] In summary, we have that \[\cos a - \cos x = 2\sin(\tfrac{x+a}{2})\sin(\tfrac{x-a}{2}),\] and hence \[\cos x - \cos a = -2\sin(\tfrac{x+a}{2})\sin(\tfrac{x-a}{2}).\]

Moving back to our limit, by applying our sum-to-product trigonometric identity we see that
\[ \begin{aligned}
\lim_{x \to a} \frac{\cos x - \cos a}{x - a} & = \lim_{x \to a} \frac{-2\sin(\tfrac{x+a}{2})\sin(\tfrac{x-a}{2})}{x - a} \\ & = -\left(\lim_{x \to a} \sin\left(\frac{x + a}{2}\right)\right)\left(\lim_{x \to a} \frac{\sin(\tfrac{x-a}{2})}{\tfrac{x-a}{2}} \right) \qquad\text{(Product Law)}
\end{aligned} \]
As \(x \to a\), \(\tfrac{x-a}{2} \to 0\), so the limit on the right is the special trigonometric limit that evaluates to \(1\).^{1} Whereas on the right, we use the continuity of the sine function to bring the limit inside.
\[ \begin{aligned}
& = -\sin\left(\lim_{x \to a} \frac{x + a}{2}\right) \cdot 1 \\ & = -\sin a
\end{aligned} \]

### Conclusion

In the solution, I have gone through the details of the calculation to a greater extent than when I first calculated the solution in my head. The solution is not advanced, but it is also not very useful to American students learning from an AP Calculus curriculum, as he/she is not likely to know the sum-to-product trigonometric identities used.^{2}

Personally, I do not *memorize* any trigonometric identities other than the laws of sine and cosine, the Pythagorean identity, and the sine and cosine angle sum and difference identities. When I need a trigonometric identity, I would usually derive it from the ones I already know.

But I think it answers the students’ curiosity in terms of whether it is possible to compute this limit without making the substitution \(h = x - a\) (which might not seem not obvious to students who are new to Calculus), and that one of the many algebraic identities students may or may not know might be key to cracking a problem like this. Although, if I had to demonstrate *only one* way to compute this limit, I would stick to the original substitution.