We are given the following expression, assuming $x$ and $y$ are small and measured in radians:

$\sin^2(x) - \sin(6y) + \frac{\sin(2x) - \sin(y)}{4}$

We will now use small-angle approximations for trigonometric functions, which are valid when $x$ and $y$ are small. The small-angle approximations are:

$\sin(x) \approx x \quad \text{and} \quad \cos(x) \approx 1$

Step-by-step evaluation:

1. First term: $\sin^2(x)$
   Since $\sin(x) \approx x$ for small $x$, we have: $\sin^2(x) \approx x^2$

2. Second term: $\sin(6y)$
   Using the small-angle approximation $\sin(6y) \approx 6y$, we get: $\sin(6y) \approx 6y$

3. Third term: $\frac{\sin(2x) - \sin(y)}{4}$
   Using the approximations $\sin(2x) \approx 2x$ and $\sin(y) \approx y$, we have: $\frac{\sin(2x) - \sin(y)}{4} \approx \frac{2x - y}{4}$

Putting it all together:

$\sin^2(x) - \sin(6y) + \frac{\sin(2x) - \sin(y)}{4} \approx x^2 - 6y + \frac{2x - y}{4}$

Now simplifying the third term:

$x^2 - 6y + \frac{2x - y}{4} = x^2 - 6y + \frac{2x}{4} - \frac{y}{4} = x^2 - 6y + \frac{x}{2} - \frac{y}{4}$

Thus, the expression is approximately:

$x^2 + \frac{x}{2} - 6y - \frac{y}{4}$

This is the simplified approximation.

Would you like further details on any specific part, or do you have any questions?

Here are 5 related questions to expand your understanding:

1. How does the small-angle approximation work for other trigonometric functions like cosine?
2. What is the importance of the small-angle approximation in physics and engineering?
3. How would the result differ if higher-order approximations were used for the sine function?
4. Could you apply a similar approximation for $\tan(x)$ in a different expression?
5. How would you evaluate this expression if $x$ and $y$ were not small?

Tip: The small-angle approximation is particularly useful in many real-world problems involving oscillations and waves where the angles involved are small enough for the approximation to hold.