KindiakMath

The Derivative of Sine

We finally are ready to prove that \sin' = \cos. All the crucial tools we developed culminates in this crucial lemma:

Lemma. \sin'(0) = 1 and \cos'(0) = 0.

Proof. Let \theta be an acute angle. By a geometric argument, considering the area of the sector subtended by \theta and two triangles,

\displaystyle \frac 12 \sin(\theta) \leq \frac 12 \theta \leq \frac 12 \cdot \frac{\sin(\theta)}{\cos(\theta)}.

Dividing by \sin(\theta)/2 on all sides,

\displaystyle 1 \leq \frac{\theta}{\sin(\theta)} \leq \frac{1}{\cos(\theta)}.

Taking reciprocals,

\displaystyle \cos(\theta) \leq \frac{\sin(\theta)}{\theta} \leq 1.

Since \cos(\theta) \to 1 as \theta \to 0, by the squeeze theorem, we have

\displaystyle \sin'(0) = \lim_{\theta \to 0} \frac{\sin(0+\theta) - \sin(0)}{\theta} = \lim_{\theta \to 0} \frac{\sin(\theta)}{\theta} = 1.

For the second identity,

\displaystyle \cos'(0) = \lim_{\theta \to 0} \frac{\cos(0+\theta) - \cos(0)}{\theta} = \lim_{\theta \to 0} \frac{\cos(\theta) - 1}{\theta}.

Applying the double angle formulae then taking \theta \to 0,

\displaystyle \frac{\cos(\theta) - 1}{\theta} = -\frac{\sin^2(\theta/2)}{\theta/2} = -\left( \frac{\sin(\theta/2)}{\theta/2} \right)^2 \cdot \frac{\theta}{2} \to -1^2 \cdot \frac 02 = 0.

Thus, \cos'(0) = 0.

Now we can prove the big result.

Theorem 1. \sin' = \cos.

Proof. Applying the addition formulae,

\sin(x+h) - \sin(x) = \sin(x) \cdot (\cos(h) - 1) + \cos(x) \cdot \sin(h).

Dividing throughout by h then taking h \to 0,

\displaystyle \begin{aligned} \frac{\sin(x+h) - \sin(x)}{h} &= \sin(x) \cdot \frac{\cos(h) - 1}{h} + \cos(x) \cdot \frac{\sin(h)}{h} \\ &\to \sin(x) \cos'(0) + \cos(x) \sin'(0) \\ &= \sin(x) \cdot 0 + \cos(x) \cdot 1 = \cos(x). \end{aligned}

In a similar manner, if we first assume that \exp'(0) = 1, we can prove one of the most important derivatives in pure and applied mathematics.

Theorem 2. \exp' = \exp.

Proof. By the definition of the derivative,

\begin{aligned} \frac{\exp(x+h) - \exp(x)}{h} &= \frac{e^{x+h}-e^x}{h} = \frac{e^x e^h - e^x}{h} = e^x \cdot \frac{e^h - 1}{h} \\ &= \exp(x) \cdot \frac{\exp(h) - \exp(0)}{h} \\ &\to \exp(x) \cdot \exp'(0) \\ &= \exp(x) \cdot 1 = \exp(x). \end{aligned}

This helps us prove that (x^r)' = rx^{r-1} for real r.

Theorem 3. For any real number r, wherever defined, (x^r)' = rx^{r-1}.

Proof. Write x^r = e^{r \ln(x)}. By the chain rule,

\displaystyle \begin{aligned} \frac{\mathrm d}{\mathrm dx}(x^r) &= \frac{\mathrm d}{\mathrm dx}( e^{r \ln(x)} ) \\ &= e^{r \ln(x)} \cdot \frac{r}{x} \\ &= x^r \cdot rx^{-1}\\ &= rx^{r-1}. \end{aligned}

Even more fascinatingly, this helps us prove that \pi is irrational. I didn’t come up with this proof. Details in the next post.

—Joel Kindiak, 27 Oct 24, 1942H

,
, , , ,

Published by

joelkindiak

Response

  1. What is an Angle? – KindiakMath

    […] the next few posts, we will finally prove that . The plan initially was to do this in a discussion in real analysis, […]

    Like

    Reply

Leave a comment