Act.trig.02 sin theta >= 1

[karush]

For $-\dfrac{\pi}{2}\le \theta \le \dfrac{\pi}{2}
\quad |\sin{\theta}\ge 1|$ is true for all and only the values of $\theta$ in which of the following sets
$a.\ \left\{-\dfrac{\pi}{2},\dfrac{\pi}{2}\right\}
\quad b.\ \left\{\dfrac{\pi}{2}\right\}
\quad c.\ \left\{\theta | -\dfrac{\pi}{2}< \theta < \dfrac{\pi}{2}\right\}
\quad d.\ \left\{\theta | -\dfrac{\pi}{2}\le \theta < \dfrac{\pi}{2} \le \right\}
\quad e.\ \textit{empty set}$

I chose b since $\sin\theta$ can only be 1 at $\dfrac{\pi}{2}$

I have seen a lot of students miss this one
not sure why

what is the symbol for an empty set? or better yet the latex code

 

[jvicens]

I would like to clarify that the statement $|\sin{\theta}\ge 1|$ is not true for all values of $\theta$ in any of the given sets. The correct answer is actually option $e$, the empty set.

The reason for this is that the absolute value of $\sin{\theta}$ can never be greater than 1. In fact, it can only take values between 0 and 1 for any value of $\theta$. This means that the given inequality $|\sin{\theta}\ge 1|$ is always false, regardless of the set of values for $\theta$.

To further explain, let's consider each option:

a. The set $\left\{-\dfrac{\pi}{2},\dfrac{\pi}{2}\right\}$ only contains two values, which are the endpoints of the given interval. Plugging these values into the inequality, we get $|\sin{\left(-\dfrac{\pi}{2}\right)}|\ge 1$ and $|\sin{\left(\dfrac{\pi}{2}\right)}|\ge 1$. But the absolute value of $\sin{\left(-\dfrac{\pi}{2}\right)}$ and $\sin{\left(\dfrac{\pi}{2}\right)}$ are both equal to 1, not greater than 1. Therefore, the inequality is still false.

b. As mentioned in the forum post, the value $\dfrac{\pi}{2}$ is the only value where $\sin{\theta}$ can be equal to 1. However, the given inequality states that $|\sin{\theta}\ge 1|$, which means that the absolute value of $\sin{\theta}$ must be greater than 1. Since this is not possible, the inequality is still false.

c. This set contains all values of $\theta$ between $-\dfrac{\pi}{2}$ and $\dfrac{\pi}{2}$, excluding the endpoints. Again, plugging in any value from this set into the inequality will result in a false statement.

d. Similar to option c, this set also contains all values of $\theta$ between $-\dfrac{\pi}{2}$ and $\dfrac{\pi}{2}$, including the endpoints. But as we have seen before, the inequality is still false for these values.

e. The empty set, denoted by $\emptyset$,