In spherical coordinates, the point $\left( 3, \frac{2 \pi}{7}, \frac{8 \pi}{5} \right)$ is equivalent to what other point, in the standard spherical coordinate representation?  Enter your answer in the form $(\rho,\theta,\phi),$ where $\rho > 0,$ $0 \le \theta < 2 \pi,$ and $0 \le \phi \le \pi.$
To find the spherical coordinates of a point $P,$ we measure the angle that $\overline{OP}$ makes with the positive $x$-axis, which is $\theta,$ and the angle that $\overline{OP}$ makes with the positive $z$-axis, which is $\phi,$ where $O$ is the origin.

[asy]
import three;

size(250);
currentprojection = perspective(6,3,2);

triple sphericaltorectangular (real rho, real theta, real phi) {
  return ((rho*Sin(phi)*Cos(theta),rho*Sin(phi)*Sin(theta),rho*Cos(phi)));
}

triple O, P;

O = (0,0,0);
P = sphericaltorectangular(1,60,45);

draw(surface(O--P--(P.x,P.y,0)--cycle),gray(0.7),nolight);
draw(O--(1,0,0),Arrow3(6));
draw(O--(0,1,0),Arrow3(6));
draw(O--(0,0,1),Arrow3(6));
draw(O--P--(P.x,P.y,0)--cycle);
draw((0,0,0.5)..sphericaltorectangular(0.5,60,45/2)..sphericaltorectangular(0.5,60,45),Arrow3(6));
draw((0.4,0,0)..sphericaltorectangular(0.4,30,90)..sphericaltorectangular(0.4,60,90),Arrow3(6));

label("$x$", (1.1,0,0));
label("$y$", (0,1.1,0));
label("$z$", (0,0,1.1));
label("$\phi$", (0.2,0.25,0.6));
label("$\theta$", (0.5,0.25,0));
label("$P$", P, N);
[/asy]

The normal ranges for $\theta$ and $\phi$ are $0 \le \theta < 2 \pi$ and $0 \le \phi \le \pi.$  Since $\phi = \frac{8 \pi}{5}$ is greater than $\pi,$ we end up wrapping past the negative $z$-axis.  Thus, $\phi$ becomes $2 \pi - \frac{8 \pi}{5} = \frac{2 \pi}{5},$ and $\theta$ becomes $\frac{2 \pi}{7} + \pi = \frac{9 \pi}{7}.$  Thus, the standard spherical coordinates are $\boxed{\left( 3, \frac{9 \pi}{7}, \frac{2 \pi}{5} \right)}.$
\le(3,\frac{9\pi}{7},\frac{2\pi}{5}\right)
Level 5
Precalculus
Problem: In spherical coordinates, the point $\left( 3, \frac{2 \pi}{7}, \frac{8 \pi}{5} \right)$ is equivalent to what other point, in the standard spherical coordinate representation?  Enter your answer in the form $(\rho,\theta,\phi),$ where $\rho > 0,$ $0 \le \theta < 2 \pi,$ and $0 \le \phi \le \pi.$
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To convert the spherical coordinates from the given representation to the standard form, we need to ensure that the azimuthal angle $\theta$ is within the range $[0, 2\pi)$ and the polar angle $\phi$ is within the range $[0, \pi]$. 

The given point is $\left( 3, \frac{2 \pi}{7}, \frac{8 \pi}{5} \right)$. Here, $\rho = 3$, which is already positive, satisfying the condition for $\rho$. 

Next, we examine the azimuthal angle $\theta = \frac{2 \pi}{7}$. This value is already within the required range of $[0, 2\pi)$, so it remains unchanged.

Now, we analyze the polar angle $\phi = \frac{8 \pi}{5}$. This value exceeds $\pi$, which violates the condition for $\phi$. To adjust $\phi$ into the standard range, we can use the property that $\phi$ can be expressed as $\phi' = 2\pi - \phi$ when $\phi > \pi$. 

Calculating this gives:
\[
\phi' = 2\pi - \frac{8\pi}{5} = \frac{10\pi}{5} - \frac{8\pi}{5} = \frac{2\pi}{5}.
\]

Since we have adjusted $\phi$, we must also adjust $\theta$. The transformation of $\phi$ from $\frac{8\pi}{5}$ to $\frac{2\pi}{5}$ implies a rotation of $\pi$ radians in the azimuthal direction. Therefore, we need to add $\pi$ to the azimuthal angle $\theta$:
\[
\theta' = \frac{2\pi}{7} + \pi = \frac{2\pi}{7} + \frac{7\pi}{7} = \frac{9\pi}{7}.
\]

Now, we have the adjusted values: $\rho = 3$, $\theta' = \frac{9\pi}{7}$, and $\phi' = \frac{2\pi}{5}$. 

Thus, the final answer in standard spherical coordinates is $\left(3, \frac{9\pi}{7}, \frac{2\pi}{5}\right)$.<|END_OF_LATENT|> The final answer is $\le(3,\frac{9\pi}{7},\frac{2\pi}{5}\right)$.