Find the smallest positive number p for which the equation `cos (p sin x) = sin (p cos x)` has a solution `x in [0,2pi]`.

Options

A. ` (pi)/(sqrt(2)) `
B. `(pi)/(2)`
C. ` (pi)/(2sqrt(2))`
D. `(3 pi)/(2 sqrt(2))`

Answer

To find the smallest positive number \( p \) for which the equation \( \cos(p \sin x) = \sin(p \cos x) \) has a solution for \( x \) in the interval \([0, 2\pi]\), we can follow these steps:

### Step 1: Rewrite the Equation
We start with the equation:
\[ \cos(p \sin x) = \sin(p \cos x) \]

### Step 2: Use the Identity
We know that \( \cos \theta = \sin\left(\frac{\pi}{2} - \theta\right) \). Therefore, we can rewrite the left-hand side:
\[ \cos(p \sin x) = \sin\left(\frac{\pi}{2} - p \sin x\right) \]
This gives us:
\[ \sin\left(\frac{\pi}{2} - p \sin x\right) = \sin(p \cos x) \]

### Step 3: Set the Angles Equal
Since the sine function is equal, we can set the angles equal to each other:
\[ \frac{\pi}{2} - p \sin x = p \cos x \]

### Step 4: Rearranging the Equation
Rearranging the equation gives:
\[ \frac{\pi}{2} = p \sin x + p \cos x \]
Factoring out \( p \):
\[ \frac{\pi}{2} = p(\sin x + \cos x) \]

### Step 5: Solve for \( p \)
Now, we can express \( p \):
\[ p = \frac{\frac{\pi}{2}}{\sin x + \cos x} \]

### Step 6: Maximize \( \sin x + \cos x \)
To find the smallest positive \( p \), we need to maximize \( \sin x + \cos x \). The maximum value of \( \sin x + \cos x \) can be found using the identity:
\[ \sin x + \cos x = \sqrt{2} \sin\left(x + \frac{\pi}{4}\right) \]
The maximum value of \( \sin\left(x + \frac{\pi}{4}\right) \) is 1, thus:
\[ \sin x + \cos x \leq \sqrt{2} \]

### Step 7: Substitute Maximum Value
Substituting this maximum value back into the equation for \( p \):
\[ p = \frac{\frac{\pi}{2}}{\sqrt{2}} = \frac{\pi}{2\sqrt{2}} = \frac{\pi \sqrt{2}}{4} \]

### Step 8: Final Result
Thus, the smallest positive number \( p \) for which the equation has a solution is:
\[ p = \frac{\pi}{2\sqrt{2}} = \frac{\pi \sqrt{2}}{4} \]