If the equation cos `(lambda "sin" theta) = "sin" (lambda "cos" theta)` has a solution in `[0, 2pi]`, then the smallest value of `lambda`, is

Options

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

Answer

To solve the equation \( \cos(\lambda \sin \theta) = \sin(\lambda \cos \theta) \) and find the smallest value of \( \lambda \) such that there is a solution in the interval \( [0, 2\pi] \), we can follow these steps:

### Step 1: Rewrite the Equation
We start with the equation:
\[ \cos(\lambda \sin \theta) = \sin(\lambda \cos \theta) \]
Using the identity \( \sin\left(\frac{\pi}{2} + x\right) = \cos(x) \), we can rewrite the sine term:
\[ \cos(\lambda \sin \theta) = \cos\left(\frac{\pi}{2} - \lambda \cos \theta\right) \]

### Step 2: Set the Angles Equal
Since the cosines are equal, we can set the angles equal to each other (considering the periodic nature of cosine):
\[ \lambda \sin \theta = \frac{\pi}{2} - \lambda \cos \theta + 2k\pi \quad \text{for some integer } k \]
or
\[ \lambda \sin \theta = -\left(\frac{\pi}{2} - \lambda \cos \theta\right) + 2k\pi \]

### Step 3: Rearranging the First Equation
From the first equation, we can rearrange it to isolate \( \lambda \):
\[ \lambda \sin \theta + \lambda \cos \theta = \frac{\pi}{2} + 2k\pi \]
Factoring out \( \lambda \):
\[ \lambda (\sin \theta + \cos \theta) = \frac{\pi}{2} + 2k\pi \]
Thus, we can express \( \lambda \) as:
\[ \lambda = \frac{\frac{\pi}{2} + 2k\pi}{\sin \theta + \cos \theta} \]

### Step 4: Finding the Minimum Value of \( \lambda \)
To find the minimum value of \( \lambda \), we need to maximize the denominator \( \sin \theta + \cos \theta \). The maximum value of \( \sin \theta + \cos \theta \) occurs when \( \theta = \frac{\pi}{4} \):
\[ \sin \theta + \cos \theta = \sqrt{2} \]
Thus, substituting this back into our expression for \( \lambda \):
\[ \lambda_{\text{min}} = \frac{\frac{\pi}{2}}{\sqrt{2}} = \frac{\pi}{2\sqrt{2}} \]

### Step 5: Simplifying the Expression
We can further simplify \( \lambda_{\text{min}} \):
\[ \lambda_{\text{min}} = \frac{\pi \sqrt{2}}{4} \]

### Conclusion
The smallest value of \( \lambda \) such that the equation has a solution in the interval \( [0, 2\pi] \) is:
\[ \lambda = \frac{\pi \sqrt{2}}{4} \]