Let's solve this step by step.

We are given the function:

$f(x) = 5 \cos x + 4 \sin x$

and we are asked to find the maximum value of $f(x)$ within the interval:

$\frac{\pi}{4} \leq x \leq \frac{3\pi}{4}$

Step 1: Rewrite the Function Using Amplitude-Phase Form

The expression $5 \cos x + 4 \sin x$ can be rewritten in the form $R \cos(x - \alpha)$ where $R$ is the amplitude and $\alpha$ is a phase shift. This can simplify finding the maximum value.

1. Calculate $R$: $R = \sqrt{5^2 + 4^2} = \sqrt{25 + 16} = \sqrt{41}$

2. Determine $\alpha$: Since $\tan \alpha = \frac{4}{5}$, we find $\alpha = \arctan\left(\frac{4}{5}\right)$.

Thus, we can rewrite the function as: $f(x) = \sqrt{41} \cos(x - \alpha)$

Step 2: Determine Maximum and Minimum Values of $f(x)$

The function $\sqrt{41} \cos(x - \alpha)$ has a range of values from $-\sqrt{41}$ to $\sqrt{41}$. Therefore, the maximum value of $f(x)$ is $\sqrt{41}$, which is approximately 6.4.

Step 3: Check the Interval Limits

Since $\cos(x - \alpha)$ oscillates between -1 and 1, the maximum value will be $\sqrt{41}$ if $x - \alpha$ is an integer multiple of $2\pi$ within the interval $\left[ \frac{\pi}{4}, \frac{3\pi}{4} \right]$.

1. At $x = \frac{\pi}{4}$: $f\left(\frac{\pi}{4}\right) = 5 \cos\left(\frac{\pi}{4}\right) + 4 \sin\left(\frac{\pi}{4}\right) = 5 \cdot \frac{\sqrt{2}}{2} + 4 \cdot \frac{\sqrt{2}}{2} = \frac{9\sqrt{2}}{2} \approx 6.36$

2. At $x = \frac{3\pi}{4}$: $f\left(\frac{3\pi}{4}\right) = 5 \cos\left(\frac{3\pi}{4}\right) + 4 \sin\left(\frac{3\pi}{4}\right) = 5 \cdot \left(-\frac{\sqrt{2}}{2}\right) + 4 \cdot \frac{\sqrt{2}}{2} = -\frac{\sqrt{2}}{2} \approx -0.707$

Conclusion

The maximum value of $f(x)$ within the interval $\frac{\pi}{4} \leq x \leq \frac{3\pi}{4}$ is approximately $\frac{9\sqrt{2}}{2} \approx 6.36$.

Would you like further details on any specific steps or have any additional questions?

Further Questions

1. How is the amplitude-phase form derived for trigonometric expressions?
2. What role does the interval play in determining the maximum value?
3. Can the maximum value change if the interval changes?
4. How would you find the minimum value in this interval?
5. What other methods can be used to find maximum and minimum values of trigonometric functions?

Tip

When working with trigonometric functions, rewriting them in terms of amplitude and phase shift can simplify finding maximum and minimum values significantly.