Which of the following functions are strictly decreasing on $(0,\frac{\pi}{2})$ ?

Question

Accepted Answer

Option 2, which is "cos 2x," is the correct answer because it is the only function among the given options that is strictly decreasing on the interval $(0, \frac{\pi}{2})$.

1. **Explanation for Option 2 (cos 2x):**
   - The function $ \cos 2x $ is a cosine function with a frequency that is doubled compared to the standard cosine function. The standard cosine function, $ \cos x $, decreases from 1 to 0 as $ x $ goes from 0 to $\frac{\pi}{2}$. When the argument is $2x$, the function $ \cos 2x $ completes its decreasing phase from 1 to -1 over the interval $ (0, \frac{\pi}) $. Therefore, within the interval $ (0, \frac{\pi}{2}) $, $ \cos 2x $ is strictly decreasing from 1 to 0.

2. **Explanation for why other options are incorrect:**

- **Option 1 (sin x):**
     - The function $ \sin x $ is not strictly decreasing on $ (0, \frac{\pi}{2}) $. In fact, it is strictly increasing on this interval, as it goes from 0 to 1.

- **Option 3 (cos 3x):**
     - The function $ \cos 3x $ is not strictly decreasing on $ (0, \frac{\pi}{2}) $. The function $ \cos 3x $ completes more than one cycle within $ (0, \frac{\pi}) $, and within $ (0, \frac{\pi}/2) $, it does not maintain a strictly decreasing behavior due to its higher frequency.

- **Option 4 (tan x):**
     - The function $ \tan x $ is not strictly decreasing on $ (0, \frac{\pi}/2) $. Instead, it is strictly increasing, as it goes from 0 to infinity.

In conclusion, among the given options, only $ \cos 2x $ is strictly decreasing on the interval $ (0, \frac{\pi}/2) $, which is why Option 2 is the correct answer.