To solve this problem, we are given $f(x) = \sin(2x + 1)$ and need to calculate the expression $\frac{1}{h} \Delta_x f(h)$, which represents a finite difference divided by $h$. Let's go through the steps.

1. Calculate $f(x + h)$: $f(x + h) = \sin(2(x + h) + 1) = \sin(2x + 2h + 1)$

2. Set up the finite difference $\Delta_x f(h)$: $\Delta_x f(h) = f(x + h) - f(x)$ Substitute $f(x + h) = \sin(2x + 2h + 1)$ and $f(x) = \sin(2x + 1)$: $\Delta_x f(h) = \sin(2x + 2h + 1) - \sin(2x + 1)$

3. Simplify using the sine difference identity: Using the identity $\sin A - \sin B = 2 \cos\left(\frac{A + B}{2}\right) \sin\left(\frac{A - B}{2}\right)$, set $A = 2x + 2h + 1$ and $B = 2x + 1$: $\Delta_x f(h) = 2 \cos\left(\frac{(2x + 2h + 1) + (2x + 1)}{2}\right) \sin\left(\frac{(2x + 2h + 1) - (2x + 1)}{2}\right)$ Simplify each part:

   • The cosine term: $\cos\left(\frac{4x + 2h + 2}{2}\right) = \cos(2x + h + 1)$
   • The sine term: $\sin\left(\frac{2h}{2}\right) = \sin(h)$ So, $\Delta_x f(h) = 2 \cos(2x + h + 1) \sin(h)$

4. Divide by $h$: $\frac{1}{h} \Delta_x f(h) = \frac{2 \cos(2x + h + 1) \sin(h)}{h}$ Using the limit $\lim_{h \to 0} \frac{\sin(h)}{h} = 1$, this expression simplifies further if $h \to 0$, but for non-zero $h$, we can leave it as: $\frac{1}{h} \Delta_x f(h) = 2 \cos(2x + h + 1) \frac{\sin(h)}{h}$

Let me know if you want more detailed steps or if you have any questions!

Here are 5 related questions to expand your understanding:

1. How would this process change if $f(x) = \cos(2x + 1)$?
2. What is the geometric interpretation of $\frac{1}{h} \Delta_x f(h)$?
3. How does the finite difference relate to the derivative as $h \to 0$?
4. Can this method be applied to other trigonometric functions, like $f(x) = \tan(x)$?
5. What happens to the result if we take $h = 0.01$ instead of letting $h \to 0$?

Tip: Understanding finite differences can provide insights into numerical approximations of derivatives, especially when analytical solutions are complex or unavailable.