Converting from cylindrical to spherical coordinates?

Question

I am supposed to convert the point $(100, -\dfrac{\pi}{6}, 50)$ from cylindrical to spherical. The $\rho$ and $\theta$ are easy ( $\sqrt{100^2+50^2}, \dfrac{\pi}{6}$ respectively) but the $\phi$ is giving me trouble. Visualizing it, I can see that $\sin\phi$ should be $\dfrac{100}{\sqrt{100^2+50^2}}$. However, the book says that $\cos\phi = \dfrac{50}{\sqrt{100^2+50^2}}$. I can see how the book is right, but I can't see why my answer is wrong (I checked numerically)

Answer 1

Since $\sin^2 \phi +\cos^2\phi =1$, then $\sin \phi =\frac{A}{\sqrt{A^2+B^2}}$ gives the equivalent angle to $\cos \phi =\frac{B}{\sqrt{A^2+B^2}}$.

For the problem of interest, $A=100$ and $B=50$ and

$$\phi = \arcsin\left(\frac{100}{\sqrt{100^2+50^2}}\right)=\arccos\left(\frac{50}{\sqrt{100^2+50^2}}\right)$$

Answer 2

Both are correct, since

$\sin\phi$ = $\dfrac{a}{\sqrt{a^2+b^2}}\,, \cos\phi = \dfrac{b}{\sqrt{a^2+b^2}}$ are different ratios of sides of the same right angled triangle.