In polar coordinates, the point $\left( -2, \frac{3 \pi}{8} \right)$ is equivalent to what other point, in the standard polar coordinate representation?  Enter your answer in the form $(r,\theta),$ where $r > 0$ and $0 \le \theta < 2 \pi.$
To obtain the point $\left( -2, \frac{3 \pi}{8} \right),$ we move counter-clockwise from the positive $x$-axis by an angle of $\frac{3 \pi}{8},$ then take the point with $r = -2$ at this angle.  Since $-2$ is negative, we end up reflecting through the origin.  Thus, we arrive at the point $\boxed{\left( 2, \frac{11 \pi}{8} \right)}.$

[asy]
unitsize(1 cm);

draw(Circle((0,0),2),red);
draw((-2.5,0)--(2.5,0));
draw((0,-2.5)--(0,2.5));
draw((0,0)--((-2)*dir(67.5)));
draw((0,0)--(2*dir(67.5)),dashed);

dot((-2)*dir(67.5));
dot(2*dir(67.6));

label("$\frac{3 \pi}{8}$", (0.5,0.3));
[/asy]
\le(2,\frac{11\pi}{8}\right)
Level 3
Precalculus
Problem: In polar coordinates, the point $\left( -2, \frac{3 \pi}{8} \right)$ is equivalent to what other point, in the standard polar coordinate representation?  Enter your answer in the form $(r,\theta),$ where $r > 0$ and $0 \le \theta < 2 \pi.$
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To convert the point $\left( -2, \frac{3 \pi}{8} \right)$ into standard polar coordinates, we first recognize that the negative radius indicates the point is in the opposite direction of the angle. The standard representation requires a positive radius. 

We can find an equivalent point by adding $\pi$ to the angle, which reflects the point across the origin. Thus, we calculate:

1. Start with the original angle: $\frac{3 \pi}{8}$.
2. Add $\pi$ to this angle: 
   \[
   \frac{3 \pi}{8} + \pi = \frac{3 \pi}{8} + \frac{8 \pi}{8} = \frac{11 \pi}{8}.
   \]

Now, we have a new angle of $\frac{11 \pi}{8}$. 

Next, we need to ensure that the radius is positive. The original radius was $-2$, so we take the absolute value to get $r = 2$. 

Finally, we check that the angle $\frac{11 \pi}{8}$ is within the required range of $0 \le \theta < 2 \pi$. Since $\frac{11 \pi}{8}$ is less than $2\pi$ (which is $\frac{16 \pi}{8}$), it satisfies this condition.

Thus, the equivalent point in standard polar coordinates is $\left(2, \frac{11 \pi}{8}\right)$.<|END_OF_LATENT|> The final answer is $\le(2,\frac{11\pi}{8}\right)$.