Change of coordinates 
Hi everyone!
Can someone explain me this exercice? My solution was $\theta\in [\frac{\pi}{4},\frac{\pi}{2}]\cup[\frac{5\pi}{4},\frac{3\pi}{2})$, which is incorrect, but I can't understand why... 
The other parts were correct.
I thought $x \leq y \Leftrightarrow r\cdot\cos(\theta)\sin(\phi) \le r\cdot \sin(\theta)\sin(\phi) \Leftrightarrow 1 \le \operatorname{tg}(\theta)$ so $\theta \in [\frac{\pi}{4},\frac{\pi}{2}]\cup[\frac{5\pi}{4},\frac{3\pi}{2})$
 A: You forgot that dividing (and multiplying) an inequality by some number requires checking the sign of the divisor (or factor), in order to find out if the inequality changes the direction.
That's because from $a \le b$ it surely does not follow $-a \le -b$, the opppsite is true: $a \le b \Leftrightarrow -a \ge -b$.
That doesn't affect the common factor $\sin(\phi)$, as that is non-negative for $0 \le \phi < \pi$. But $\cos(\theta)$ changes signs at $\theta=\frac{\pi}2,\frac{3\pi}2$. That means when you divide by it to get the inequality involving $\tan(\theta)$ you need to consider 2 cases:
1) $\cos(\theta) \ge 0$, or $\theta \in [0,{\pi \over 2}] \cup [{3\pi \over 2},{2\pi})$. Your correct solution set for $\tan(\theta) \ge 1$ needs then be restricted to the given interval of $\theta$, which gives $I_1(\theta)=[{\pi \over 4},{\pi \over 2}]$.
2) $\cos(\theta) < 0$, or $\theta \in ({\pi \over 2}, {3\pi \over 2})$. In that interval, then the inequality becomes $\tan(\theta) \le 1$. The solution set of this, restricted to $({\pi \over 2}, {3\pi \over 2})$ is $I_2(\theta)=({\pi \over 2},{5\pi \over 4}]$.
Finally, the possible set for $\theta$ is $I_1(\theta) \cup I_2(\theta) = [{\pi \over 4},{5\pi \over 4}]$, as given in the book.
So the main thing to remember is that multiplying/dividing an inequality requires extra steps to get the correct direction of the inequality!
