To solve this problem, we will go through each part step-by-step.

Given Information:

• $f(x) = \cos x$
• Point of linearization: $a = \frac{\pi}{2}$
• The value we want to approximate: $f(1.53)$

Steps:

a. Write the equation of the line that represents the linear approximation at $a = \frac{\pi}{2}$.

1. Find $f(a)$: $f\left(\frac{\pi}{2}\right) = \cos\left(\frac{\pi}{2}\right) = 0$

2. Find $f'(x)$: Since $f(x) = \cos x$, its derivative $f'(x) = -\sin x$.

3. Evaluate $f'(a)$: $f'\left(\frac{\pi}{2}\right) = -\sin\left(\frac{\pi}{2}\right) = -1$

4. Form the linear approximation $L(x)$: The linear approximation formula is: $L(x) = f(a) + f'(a)(x - a)$ Substituting in the values we found: $L(x) = 0 + (-1)(x - \frac{\pi}{2})$ Simplifying, we get: $L(x) = -\left(x - \frac{\pi}{2}\right) = \frac{\pi}{2} - x$

So, the equation of the line is: $L(x) = \frac{\pi}{2} - x$

b. Use the linear approximation to estimate $f(1.53)$.

1. Substitute $x = 1.53$ into the linear approximation $L(x)$: $L(1.53) = \frac{\pi}{2} - 1.53$

2. Using $\frac{\pi}{2} \approx 1.5708$: $L(1.53) = 1.5708 - 1.53 = 0.0408$

So, the linear approximation for $f(1.53)$ is approximately 0.0408.

c. Compute the percent error in the approximation.

1. Calculate the exact value of $f(1.53)$: $f(1.53) = \cos(1.53)$ Using a calculator, $\cos(1.53) \approx 0.0424$.

2. Calculate the percent error using the formula: $\text{Percent Error} = 100 \cdot \frac{|\text{approximation} - \text{exact}|}{|\text{exact}|}$ Substitute the values: $\text{Percent Error} = 100 \cdot \frac{|0.0408 - 0.0424|}{|0.0424|}$

3. Simplifying further: $\text{Percent Error} = 100 \cdot \frac{0.0016}{0.0424} \approx 3.77\%$

Final Answers:

• a. The equation of the line is $L(x) = \frac{\pi}{2} - x$.
• b. The linear approximation of $f(1.53)$ is approximately 0.0408.
• c. The percent error in the approximation is approximately 3.77%.

Would you like further details or have any questions about this solution?

Related Questions:

1. What is the linear approximation method, and why is it useful?
2. How do you find the derivative of trigonometric functions?
3. Why does the percent error matter in approximations?
4. How would the approximation change if we chose a different point $a$ for linearization?
5. Can the linear approximation method be used for non-trigonometric functions?

Tip:

When using linear approximation, remember that accuracy decreases as you move farther from the point of tangency, making the approximation less reliable for points further from $a$.