Allowing $x$ to be a real number, what is the largest value that can be obtained by the function $25\sin(4x)-60\cos(4x)?$
1. Consider the function \( f(x) = 25\sin(4x) - 60\cos(4x) \).
2. To find the maximum value, we can use calculus by taking the derivative and setting it to zero. First, compute the derivative:
   \[
   f'(x) = 100\cos(4x) + 240\sin(4x)
   \]
3. Set the derivative equal to zero to find the critical points:
   \[
   100\cos(4x) + 240\sin(4x) = 0
   \]
4. Divide by \(\cos(4x)\) (assuming \(\cos(4x) \neq 0\)):
   \[
   100 + 240\tan(4x) = 0
   \]
5. Solve for \(\tan(4x)\):
   \[
   \tan(4x) = -\frac{5}{12}
   \]
6. Using the identity \(\tan^2(4x) + 1 = \sec^2(4x)\), we can find \(\sin(4x)\) and \(\cos(4x)\). First, find \(\sec(4x)\):
   \[
   \sec^2(4x) = 1 + \left(-\frac{5}{12}\right)^2 = 1 + \frac{25}{144} = \frac{169}{144}
   \]
   \[
   \sec(4x) = \pm \frac{13}{12}
   \]
7. Therefore, \(\cos(4x) = \pm \frac{12}{13}\). Since \(\tan(4x) = -\frac{5}{12}\), we are in the second or fourth quadrant where \(\cos(4x)\) is negative:
   \[
   \cos(4x) = -\frac{12}{13}
   \]
8. Using \(\sin^2(4x) + \cos^2(4x) = 1\):
   \[
   \sin^2(4x) = 1 - \left(-\frac{12}{13}\right)^2 = 1 - \frac{144}{169} = \frac{25}{169}
   \]
   \[
   \sin(4x) = \pm \frac{5}{13}
   \]
9. To maximize the original function, we need to choose the correct signs for \(\sin(4x)\) and \(\cos(4x)\). Since \(\cos(4x)\) is negative and \(\sin(4x)\) is positive in the second quadrant:
   \[
   \sin(4x) = \frac{5}{13}, \quad \cos(4x) = -\frac{12}{13}
   \]
10. Substitute these values back into the original function:
    \[
    f(x) = 25\left(\frac{5}{13}\right) - 60\left(-\frac{12}{13}\right)
    \]
    \[
    f(x) = \frac{125}{13} + \frac{720}{13} = \frac{845}{13} = 65
    \]

The final answer is \(\boxed{65}\).