### Instruction: 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 $0 \le \phi \le \pi.$.### Response: To convert from spherical to Cartesian coordinates, I can use the formulas $x = \rho \sin \phi \cos \theta,$ $y = \rho \sin \phi \sin \theta,$ and $z = \rho \cos \phi.$
Plugging in the given values, I get $x = 3 \sin \frac{8 \pi}{5} \cos \frac{2 \pi}{7},$ $y = 3 \sin \frac{8 \pi}{5} \sin \frac{2 \pi}{7},$ and $z = 3 \cos \frac{8 \pi}{5}.$
I can simplify these expressions using the fact that $\sin \frac{8 \pi}{5} = - \sin \frac{2 \pi}{5}$ and $\cos \frac{8 \pi}{5} = \cos \frac{2 \pi}{5}.$
So I get $x = -3 \sin \frac{2 \pi}{5} \cos \frac{2 \pi}{7},$ $y = -3 \sin \frac{2 \pi}{5} \sin \frac{2 \pi}{7},$ and $z = 3 \cos \frac{2 \pi}{5}.$
To convert back to spherical coordinates, I can use the formulas $\rho = \sqrt{x^2 + y^2 + z^2},$ $\theta = \tan^{-1} \frac{y}{x},$ and $\phi = \cos^{-1} \frac{z}{\rho}.$
To find $\rho,$ I can square each of the Cartesian coordinates and add them up.
I get $\rho^2 = 9 \sin^2 \frac{2 \pi}{5} \cos^2 \frac{2 \pi}{7} + 9 \sin^2 \frac{2 \pi}{5} \sin^2 \frac{2 \pi}{7} + 9 \cos^2 \frac{2 \pi}{5}.$
I can factor out the $9$ and use the identity $\sin^2 \theta + \cos^2 \theta = 1$ to simplify this expression.
I get $\rho^2 = 9 \sin^2 \frac{2 \pi}{5} (\cos^2 \frac{2 \pi}{7} + \sin^2 \frac{2 \pi}{7}) + 9 \cos^2 \frac{2 \pi}{5}.$
This becomes $\rho^2 = 9 \sin^2 \frac{2 \pi}{5} + 9 \cos^2 \frac{2 \pi}{5}.$
I can apply the same identity again and get $\rho^2 = 9.$
Taking the square root, I get $\rho = 3.$
To find $\theta,$ I can use the inverse tangent function and the ratio of the $y$ and $x$ coordinates.
I get $\theta = \tan^{-1} \frac{-3 \sin \frac{2 \pi}{5} \sin \frac{2 \pi}{7}}{-3 \sin \frac{2 \pi}{5} \cos \frac{2 \pi}{7}}.$
I can cancel out the $-3$ and the $\sin \frac{2 \pi}{5}$ in the numerator and denominator, and get $\theta = \tan^{-1} \frac{\sin \frac{2 \pi}{7}}{\cos \frac{2 \pi}{7}}.$
This is just the tangent of the angle $\frac{2 \pi}{7},$ so I get $\theta = \frac{2 \pi}{7}.$
To find $\phi,$ I can use the inverse cosine function and the ratio of the $z$ and $\rho$ coordinates.
I get $\phi = \cos^{-1} \frac{3 \cos \frac{2 \pi}{5}}{3}.$
I can cancel out the $3$ in the numerator and denominator, and get $\phi = \cos^{-1} \cos \frac{2 \pi}{5}.$
This is just the angle $\frac{2 \pi}{5},$ so I get $\phi = \frac{2 \pi}{5}.$