To solve for the equation of the tangent line to the curve $y = \sin(\sin(x))$ at the given point $(2\pi, 0)$, we follow these steps:

1. Find the derivative of $y = \sin(\sin(x))$

We need to compute the derivative $\frac{dy}{dx}$. Using the chain rule, we get: $\frac{dy}{dx} = \cos(\sin(x)) \cdot \cos(x)$ This derivative will give us the slope of the tangent line at any point $x$.

2. Evaluate the derivative at $x = 2\pi$

Now, substitute $x = 2\pi$ into the derivative to find the slope of the tangent line at that point: $\frac{dy}{dx} \Big|_{x=2\pi} = \cos(\sin(2\pi)) \cdot \cos(2\pi)$ Since $\sin(2\pi) = 0$ and $\cos(2\pi) = 1$, we have: $\frac{dy}{dx} \Big|_{x=2\pi} = \cos(0) \cdot 1 = 1$ So, the slope of the tangent line at $(2\pi, 0)$ is 1.

3. Use the point-slope form to find the equation of the tangent line

The equation of a line is given by the point-slope form: $y - y_1 = m(x - x_1)$ where $m$ is the slope and $(x_1, y_1)$ is the point of tangency. Substituting $m = 1$, $x_1 = 2\pi$, and $y_1 = 0$, we get: $y - 0 = 1(x - 2\pi)$ which simplifies to: $y = x - 2\pi$

Final Answer:

The equation of the tangent line is $y = x - 2\pi$.

Would you like more details or have any other questions?

Here are 5 related questions to expand your understanding:

1. How would the derivative change if the function were $y = \cos(\sin(x))$?
2. Can the tangent line ever be horizontal for the function $y = \sin(\sin(x))$?
3. What is the second derivative of $y = \sin(\sin(x))$, and what does it represent geometrically?
4. How does the chain rule apply to more complicated compositions of trigonometric functions?
5. How do you find the tangent line to a parametric curve?

Tip: Always remember to check the values of trigonometric functions at key points like $0, \pi, 2\pi$, as they can often simplify your calculations!