To use linear approximation to estimate $\cos(28.5^\circ)$, we can utilize the formula for linear approximation:

$f(x) \approx f(a) + f'(a)(x - a)$

Here, $f(x) = \cos(x)$, and we need to choose a point $a$ that is close to $28.5^\circ$ for which we can easily calculate $f(a)$ and $f'(a)$.

1. Choose $a$: A suitable choice is $a = 30^\circ$ because it is close to $28.5^\circ$.

2. Calculate $f(a)$: $f(30^\circ) = \cos(30^\circ) = \frac{\sqrt{3}}{2} \approx 0.8660$

3. Calculate $f'(x)$: The derivative $f'(x) = -\sin(x)$. Thus: $f'(30^\circ) = -\sin(30^\circ) = -\frac{1}{2} = -0.5$

4. Apply the linear approximation: $f(28.5^\circ) \approx f(30^\circ) + f'(30^\circ)(28.5^\circ - 30^\circ)$ $\approx 0.8660 + (-0.5)(28.5^\circ - 30^\circ)$ $= 0.8660 + (-0.5)(-1.5) = 0.8660 + 0.75 = 1.6160$

Since this value is not valid for cosine (it exceeds 1), let's adjust our approach for the linear approximation:

Calculating $28.5^\circ$ in radians: $28.5^\circ = \frac{28.5 \times \pi}{180} \approx 0.4974 \text{ radians}$

Using a more accurate approximation:

1. Using $a = 0.5236$ (30° in radians): $f(0.5236) \approx 0.8660$ $f'(0.5236) = -0.5$ $f(28.5^\circ) \approx 0.8660 - 0.5(0.4974 - 0.5236)$ $= 0.8660 - 0.5(-0.0262) = 0.8660 + 0.0131 \approx 0.8791$

So, the linear approximation of $\cos(28.5^\circ)$ is approximately $0.8791$.

Final Answer: $\cos(28.5^\circ) \approx 0.8791$ (rounded to four decimal places).

Would you like more details on this method or have any other questions? Here are some related questions you might consider:

1. What is the exact value of $\cos(28.5^\circ)$?
2. How does the accuracy of linear approximation change with the choice of $a$?
3. What other functions can be approximated using differentials?
4. Can you explain how to derive the linear approximation formula?
5. What are some practical applications of linear approximation in real life?

Tip: When choosing $a$ for linear approximation, pick a value where the function is easy to compute.