Let us see how to calculate the derivative of the sine and cosine functions, using the limit of the difference quotient and fundamental trigonometric identities. We demonstrate step by step that the derivative of \( \sin(x) \) is \( \cos(x) \) and that of \( \cos(x) \) is \( -\sin(x) \), justifying each step in a clear and formal manner.
Table of Contents
Derivative of the Sine Function
We calculate the derivative of the function \( f(x) = \sin(x) \) as the limit of the difference quotient:
\[ f'(x_0) = \lim_{h \to 0} \frac{f(x_0 + h) - f(x_0)}{h} \] \[ = \lim_{h \to 0} \frac{\sin(x_0 + h) - \sin(x_0)}{h} \]
We use the trigonometric identity for the difference of sines:
\[ \sin(x_0 + h) - \sin(x_0) = 2 \sin\left(\frac{h}{2}\right) \cos\left(x_0 + \frac{h}{2}\right) \]
Substituting this identity into the difference quotient, we obtain:
\begin{align} \frac{\sin(x_0 + h) - \sin(x_0)}{h} &= \frac{2 \sin\left(\frac{h}{2}\right) \cos\left(x_0 + \frac{h}{2}\right)}{h} \\ &= \frac{\sin\left(\frac{h}{2}\right) \cos\left(x_0 + \frac{h}{2}\right)}{\frac{h}{2}} \end{align}
We know that:
\[ \lim_{h \to 0} \frac{\sin(\frac{h}{2})}{\frac{h}{2}} = 1 \quad \text{and} \quad \lim_{h \to 0} \cos\left(x_0 + \frac{h}{2}\right) = \cos(x_0). \]
Therefore:
\[ \lim_{h \to 0} \frac{\sin(x_0 + h) - \sin(x_0)}{h} = \cos(x_0) \]
The derivative of the function \( \sin(x) \) is therefore:
\[ f'(x) = \cos(x) \quad , \quad \forall x \in \mathbb{R} \]
Derivative of the Cosine Function
Now we calculate the derivative of the function \( g(x) = \cos(x) \) as the limit of the difference quotient:
\[ g'(x_0) = \lim_{h \to 0} \frac{g(x_0 + h) - g(x_0)}{h} \] \[ = \lim_{h \to 0} \frac{\cos(x_0 + h) - \cos(x_0)}{h} \]
We use the trigonometric identity for the difference of cosines:
\[ \cos(x_0 + h) - \cos(x_0) = -2 \sin\left(x_0 + \frac{h}{2}\right) \sin\left(\frac{h}{2}\right) \]
Substituting this identity into the difference quotient, we obtain:
\begin{align} \frac{\cos(x_0 + h) - \cos(x_0)}{h} &= \frac{-2 \sin\left(x_0 + \frac{h}{2}\right) \sin\left(\frac{h}{2}\right)}{h} \\ &= \frac{-\sin\left(x_0 + \frac{h}{2}\right) \sin\left(\frac{h}{2}\right)}{\frac{h}{2}}\end{align}
We know that:
\[ \lim_{h \to 0} \frac{\sin(\frac{h}{2})}{\frac{h}{2}} = 1 \quad \text{and} \quad \lim_{h \to 0} \sin\left(x_0 + \frac{h}{2}\right) = \sin(x_0). \]
Therefore:
\[ \lim_{h \to 0} \frac{\cos(x_0 + h) - \cos(x_0)}{h} = -\sin(x_0) \]
The derivative of the function \( g(x) = \cos(x) \) is therefore:
\[ g'(x) = -\sin(x) \quad , \quad \forall x \in \mathbb{R} \]