If `sin(pi cos theta)=cos(pi sin theta)`, then `sin 2theta=`

Options

A. x may be a multiple of `pi`
B. x can not be an even multiple of `pi`
C. z can be a multiple of `pi`
D. y can be a multiple of `pi//2`

Answer

To solve the equation \( \sin(\pi \cos \theta) = \cos(\pi \sin \theta) \), we will follow these steps:

### Step 1: Rewrite the Cosine Function
We know that \( \cos\left(\frac{\pi}{2} - x\right) = \sin(x) \). Thus, we can rewrite the right-hand side of the equation:
\[ \cos(\pi \sin \theta) = \sin\left(\frac{\pi}{2} - \pi \sin \theta\right) \]
So, we can rewrite the equation as:
\[ \sin(\pi \cos \theta) = \sin\left(\frac{\pi}{2} - \pi \sin \theta\right) \]

### Step 2: Set Up the Equation
Since the sine function is periodic, we can set up the following equations:
\[ \pi \cos \theta = \frac{\pi}{2} - \pi \sin \theta + 2n\pi \quad \text{or} \quad \pi \cos \theta = \pi - \left(\frac{\pi}{2} - \pi \sin \theta\right) + 2n\pi \]
where \( n \) is any integer.

### Step 3: Solve the First Equation
From the first equation:
\[ \pi \cos \theta + \pi \sin \theta = \frac{\pi}{2} + 2n\pi \]
Dividing through by \( \pi \):
\[ \cos \theta + \sin \theta = \frac{1}{2} + 2n \]

### Step 4: Solve the Second Equation
From the second equation:
\[ \pi \cos \theta + \pi \sin \theta = \frac{3\pi}{2} + 2n\pi \]
Dividing through by \( \pi \):
\[ \cos \theta + \sin \theta = \frac{3}{2} + 2n \]

### Step 5: Analyze the Possible Cases
Now we will analyze both cases:
1. **Case 1**: \( \cos \theta + \sin \theta = \frac{1}{2} + 2n \)
2. **Case 2**: \( \cos \theta + \sin \theta = \frac{3}{2} + 2n \)

### Step 6: Square Both Sides
For both cases, we can square both sides to eliminate the sine and cosine terms:
\[ (\cos \theta + \sin \theta)^2 = \left(\frac{1}{2} + 2n\right)^2 \quad \text{and} \quad (\cos \theta + \sin \theta)^2 = \left(\frac{3}{2} + 2n\right)^2 \]
Using the identity \( \cos^2 \theta + \sin^2 \theta + 2\sin \theta \cos \theta = 1 + \sin 2\theta \):
\[ 1 + \sin 2\theta = \left(\frac{1}{2} + 2n\right)^2 \quad \text{and} \quad 1 + \sin 2\theta = \left(\frac{3}{2} + 2n\right)^2 \]

### Step 7: Solve for \( \sin 2\theta \)
From the first case:
\[ 1 + \sin 2\theta = \frac{1}{4} + 2n + 2n^2 \]
Thus,
\[ \sin 2\theta = \frac{1}{4} - 1 + 2n + 2n^2 = 2n + 2n^2 - \frac{3}{4} \]

From the second case:
\[ 1 + \sin 2\theta = \frac{9}{4} + 3n + 2n^2 \]
Thus,
\[ \sin 2\theta = 3n + 2n^2 - \frac{1}{4} \]

### Step 8: Conclusion
By solving both cases, we find that:
\[ \sin 2\theta = \pm \frac{3}{4} \]