Suppose that $x$ is measured in radians. Find the maximum value of \[\frac{\sin2x+\sin4x+\sin6x}{\cos2x+\cos4x+\cos6x}\] for $0\le x\le \frac{\pi}{16}$
1. We need to find the maximum value of the expression 
   \[
   \frac{\sin 2x + \sin 4x + \sin 6x}{\cos 2x + \cos 4x + \cos 6x}
   \]
   for \(0 \le x \le \frac{\pi}{16}\).

2. First, observe that the numerator and the denominator are sums of sine and cosine functions, respectively. We can use trigonometric identities to simplify the expression.

3. Notice that the maximum value of each sine function occurs when its argument is \(\frac{\pi}{2}\) plus any multiple of \(2\pi\), and the minimum value of each cosine function occurs when its argument is \(\pi\) plus any multiple of \(2\pi\). However, since \(x\) is in the range \([0, \frac{\pi}{16}]\), we need to evaluate the expression at the boundary of this interval.

4. Evaluate the expression at \(x = \frac{\pi}{16}\):
   \[
   \frac{\sin 2\left(\frac{\pi}{16}\right) + \sin 4\left(\frac{\pi}{16}\right) + \sin 6\left(\frac{\pi}{16}\right)}{\cos 2\left(\frac{\pi}{16}\right) + \cos 4\left(\frac{\pi}{16}\right) + \cos 6\left(\frac{\pi}{16}\right)}
   \]
   Simplify the arguments:
   \[
   = \frac{\sin \frac{\pi}{8} + \sin \frac{\pi}{4} + \sin \frac{3\pi}{8}}{\cos \frac{\pi}{8} + \cos \frac{\pi}{4} + \cos \frac{3\pi}{8}}
   \]

5. Use the fact that \(\sin \theta = \cos (\frac{\pi}{2} - \theta)\) to rewrite the numerator and the denominator:
   \[
   = \frac{\sin \frac{\pi}{8} + \sin \frac{2\pi}{8} + \sin \frac{3\pi}{8}}{\cos \frac{\pi}{8} + \cos \frac{2\pi}{8} + \cos \frac{3\pi}{8}}
   \]

6. Notice that \(\sin \frac{3\pi}{8} = \cos \frac{\pi}{8}\), \(\sin \frac{2\pi}{8} = \cos \frac{\pi}{4}\), and \(\sin \frac{\pi}{8} = \cos \frac{3\pi}{8}\). Therefore, the numerator and the denominator are equal:
   \[
   = \frac{\cos \frac{\pi}{8} + \cos \frac{\pi}{4} + \cos \frac{3\pi}{8}}{\cos \frac{\pi}{8} + \cos \frac{\pi}{4} + \cos \frac{3\pi}{8}} = 1
   \]

7. Thus, the maximum value of the given expression is 1.

The final answer is \(\boxed{1}\).