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 \( \cos(\lambda \sin \theta) = \sin(\lambda \cos \theta) \) for the smallest value of \( \lambda \) such that this equation has a solution in the interval \( [0, 2\pi] \), we can follow these steps:

### Step 1: Rewrite the equation
The given equation is:
\[ \cos(\lambda \sin \theta) = \sin(\lambda \cos \theta) \]
We can use the identity that relates sine and cosine:
\[ \sin(x) = \cos\left(\frac{\pi}{2} - x\right) \]
Thus, we can rewrite the equation as:
\[ \cos(\lambda \sin \theta) = \cos\left(\frac{\pi}{2} - \lambda \cos \theta\right) \]

### Step 2: Set the arguments equal
Since the cosine function is periodic, we can set the arguments equal to each other:
\[ \lambda \sin \theta = \frac{\pi}{2} - \lambda \cos \theta + 2k\pi \quad (k \in \mathbb{Z}) \]
For simplicity, we will consider the case when \( k = 0 \):
\[ \lambda \sin \theta + \lambda \cos \theta = \frac{\pi}{2} \]

### Step 3: Factor out \( \lambda \)
We can factor \( \lambda \) out from the left-hand side:
\[ \lambda (\sin \theta + \cos \theta) = \frac{\pi}{2} \]
Thus, we can express \( \lambda \) as:
\[ \lambda = \frac{\frac{\pi}{2}}{\sin \theta + \cos \theta} \]

### Step 4: Find the maximum of \( \sin \theta + \cos \theta \)
To minimize \( \lambda \), we need to maximize \( \sin \theta + \cos \theta \). The maximum value of \( \sin \theta + \cos \theta \) can be found using the identity:
\[ \sin \theta + \cos \theta = \sqrt{2} \sin\left(\theta + \frac{\pi}{4}\right) \]
The maximum value of \( \sin\left(\theta + \frac{\pi}{4}\right) \) is 1, so:
\[ \max(\sin \theta + \cos \theta) = \sqrt{2} \]

### Step 5: Substitute back to find \( \lambda \)
Now substituting back into the expression for \( \lambda \):
\[ \lambda = \frac{\frac{\pi}{2}}{\sqrt{2}} = \frac{\pi}{2\sqrt{2}} = \frac{\pi \sqrt{2}}{4} \]

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