How can I prove the derivative of $\sin(x)$ = $\cos(x)$ using two x values a and b – where b approaches a.
The first step here would be:
$$\lim_{a\to b}\frac{\sin (b) – \sin(a)}{b-a}$$
Derivative of sin(x) = cos(x) from first principles without using (a + h)
calculustrigonometry
Related Solutions
When we have a question of calculating the derivative via first principles then it means that the idea is to drill down the definition of derivative via actual examples. It also signifies that the student is beginning to learn differential calculus. It is therefore much better to use techniques which rely on standard limits and don't rely on advanced theorems of differential calculus (like Taylor's series on L'Hospital's rule).
We will make use of the following standard limits $$\lim_{h \to 0}\frac{e^{h} - 1}{h} = 1, \,\,\lim_{h \to 0}\frac{\sin h}{h} = 1$$ We have $f(x) = e^{\cos x}$ and by definition of derivative $$\begin{aligned}f'(x)\, &= \lim_{h \to 0}\frac{f(x + h) - f(x)}{h}\\ &= \lim_{h \to 0}\frac{e^{\cos(x + h)} - e^{\cos x}}{h}\\ &= \lim_{h \to 0}\frac{e^{\cos x}\{e^{\cos(x + h) - \cos x} - 1\}}{h}\\ &= e^{\cos x}\lim_{h \to 0}\frac{e^{\cos(x + h) - \cos x} - 1}{\cos(x + h) - \cos x}\cdot\frac{\cos(x + h) - \cos x}{h}\\ &= e^{\cos x}\lim_{t \to 0}\frac{e^{t} - 1}{t}\cdot\lim_{h \to 0}\frac{\cos(x + h) - \cos x}{h}\\ &= e^{\cos x}\cdot 1\cdot\lim_{h \to 0}\dfrac{-2\sin\left(x + \dfrac{h}{2}\right)\sin\left(\dfrac{h}{2}\right)}{h}\\ &= -e^{\cos x}\cdot\lim_{h \to 0}\sin\left(x + \frac{h}{2}\right)\cdot\lim_{h \to 0}\frac{\sin(h/2)}{h/2}\\ &= -e^{\cos x}\cdot\sin x\cdot 1 = -e^{\cos x}\sin x\end{aligned}$$ In the above derivation we have made the substitution $t = \cos(x + h) - \cos x$ and $t \to 0$ as $h \to 0$.
Here is a different approach that uses the integral definition of the arcsine function. We will deduce the limit of interest, $\lim_{h\to 0}\frac{\sin(h)}{h}=1$, without appeal to geometry or differential calculus. (Note that $\cos(h)-1=-2\sin^2(h/2)$)
Instead, we only rely on elementary analysis of continuous functions and their inverses along with simple properties of the Riemann integral. To that end, we now proceed.
We define the sine function, $\sin(x)$, as the inverse function of the function $f(x)$ given by
$$\bbox[5px,border:2px solid #C0A000]{f(x)=\int_0^x \frac{1}{\sqrt{1-t^2}}\,dt }\tag 1$$
for $|x|< 1$.
NOTE: It can be shown that the sine function defined as the inverse of $f(x)$ given in $(1)$ has all of the familiar properties that characterize the circular function $\sin(x)$.
It is straightforward to show that since $\frac{1}{\sqrt{1-t^2}}$ is positive and continuous for $t\in (-1,1)$, $f(x)$ is continuous and strictly increasing for $x\in (-1,1)$ with $\displaystyle\lim_{x\to 0}f(x)=f(0)=0$.
Therefore, since $f$ is continuous and strictly increasing, its inverse function, $\sin(x)$, exists and is also continuous and strictly increasing with $\displaystyle \lim_{x\to 0}\sin(x)=\sin(0)=0$.
From $(1)$, we have the bounds (SEE HERE)
$$\bbox[5px,border:2px solid #C0A000]{1 \le \frac{f(x)}x\le \frac{1}{\sqrt{1-x^2}}} \tag 2$$
for $x\in (-1,1)$, whence applying the squeeze theorem to $(2)$ yields
$$\lim_{x\to 0}\frac{f(x)}{x}=1 \tag 3$$
Finally, let $y=f(x)$ so that $x=\sin(y)$. As $x\to 0$, $y\to 0$ and we can write $(3)$ as
$$\lim_{y\to 0}\frac{y}{\sin(y)}=1$$
from which we have
$$\bbox[5px,border:2px solid #C0A000]{\lim_{y\to 0}\frac{\sin(y)}{y}=1}$$
as was to be shown!
NOTE:
We can deduce the following set of useful inequalities from $(2)$. We let $x=\sin(\theta)$ and restrict $x$ so that $x\in [0,1)$. In addition, we define new functions, $\cos(\theta)=\sqrt{1-\sin^2(\theta)}$ and $\tan(\theta)=\sin(\theta)/\cos(\theta)$.
Then, we have from $(2)$
$$\bbox[5px,border:2px solid #C0A000]{y\cos(y)\le \sin(y)\le y\le \tan(y)} $$
which are the familiar inequalities often introduced in an introductory geometry or trigonometry course.
Best Answer
I'm sorry but I'll use a notation that is more convenient for me to write in. Hope that's ok. The connection of course would be that $b=x$ and $a=x_0$
Consider the following: for every $x\neq x_o$ we have $$ \frac{\sin (x)-\sin \left(x_{0}\right)}{x-x_{0}}=\frac{2 \sin \left(\frac{x-x_{0}}{2}\right) \cdot \cos \left(\frac{x+x_{0}}{2}\right)}{x-x_{0}}=\frac{\sin \left(\frac{x-x_{0}}{2}\right)}{\frac{x-x_{0}}{2}} \cdot \cos \left(\frac{x+x_{0}}{2}\right) $$
By the trig identity $ \sin \alpha+\sin \beta=2 \sin \left(\frac{\alpha+\beta}{2}\right) \cos \left(\frac{\alpha-\beta}{2}\right) $. Now we substitute $y=\frac{x-x_{0}}{2}$, and see that: $$ \lim _{x \rightarrow x_{0}} \frac{\sin \left(\frac{x-x_{0}}{2}\right)}{\frac{x-x_{0}}{2}}=\lim _{y \rightarrow 0} \frac{\sin y}{y}=1 $$ And: $$ \lim _{x \rightarrow x_{0}} \cos \left(\frac{x+x_{0}}{2}\right)=\cos x_{0} $$ After manipulation we arrived at limit of product, since both limits exist it'll be equal to the product of the limits, so: $$ \lim _{x \rightarrow x_{0}} \frac{\sin (x)-\sin \left(x_{0}\right)}{x-x_{0}}=\lim _{x \rightarrow x_{0}} \frac{\sin \left(\frac{x-x_{0}}{2}\right)}{\frac{x-x_{0}}{2}} \cdot \cos \left(\frac{x+x_{0}}{2}\right)=1 \cdot \cos x_{0}=\cos{(x_0)} $$
As required.