Derivative of sin(x) - Proof and Explanation

Proof

Explanation

1. The proof begins by stating the definition of the derivative of a real function at a point. In this case, it’s the derivative of sine with respect to $x$, which is the limit as $h$ approaches $0$ of $\frac{\sin (x+h)-\sin (x)}{h}$.

2. The next step uses the trigonometric identity for the sine of a sum: $\sin (A+B)=\sin (A)\cos (B)+\cos (A)\sin (B)$. Here, $A$ is $x$ and $B$ is $h$. Applying this identity to $\sin (x+h)$, we get: $\sin (x)\cos (h)+\sin (h)\cos (x)$.

3. The numerator is then rearranged by collecting the terms containing $\sin (x)$. Specifically, $\sin (x)$ is factored out from the terms involving it and $\sin (x)\cos (h)-\sin (x)$ is factored to $\sin (x)(\cos (h)-1)$, and $\sin (h)\cos (x)$ is left as is. The denominator $h$ remains unchanged.

4. The limit is split into two parts using the sum rule for limits. This rule states that the limit of a sum is equal to the sum of the limits, provided both limits exist. So we now have two limits: one for $\frac{\sin (x)(\cos (h)-1)}{h}$ and another for $\frac{\sin (h)\cos (x)}{h}$.

5. We can evaluate each of these limits separately. The limit of $\frac{\sin (h)}{h}$ as $h$ approaches $0$ is equal to $1$ (this is a standard limit). The limit of $\frac{\cos (h)-1}{h}$ as $h$ approaches $0$ is equal to $0$ (this is another standard limit). When we multiply these limits by $\sin (x)$ and $\cos (x)$ respectively, we get $\sin (x)·0$ and $1·\cos (x)$.

6. Adding these together as per the sum rule for limits, we get $0+\cos (x)$, which simplifies to $\cos (x)$.

QED: Therefore, the derivative of $\sin (x)$ with respect to $x$ is $\boxed{\cos (x)}$.