Find minimum value of `2sin^(2)theta+32cosec^(2)theta`.

Answer

To find the minimum value of the expression \(2\sin^2\theta + 32\csc^2\theta\), we can follow these steps:

### Step 1: Rewrite the expression
We start with the expression:
\[ y = 2\sin^2\theta + 32\csc^2\theta \]
We know that \(\csc^2\theta = \frac{1}{\sin^2\theta}\), so we can rewrite the expression as:
\[ y = 2\sin^2\theta + 32\frac{1}{\sin^2\theta} \]

### Step 2: Let \(x = \sin^2\theta\)
Let \(x = \sin^2\theta\). Then the expression becomes:
\[ y = 2x + \frac{32}{x} \]
where \(0 < x \leq 1\).

### Step 3: Find the derivative
To find the minimum value, we need to take the derivative of \(y\) with respect to \(x\):
\[ \frac{dy}{dx} = 2 - \frac{32}{x^2} \]

### Step 4: Set the derivative to zero
Set the derivative equal to zero to find critical points:
\[ 2 - \frac{32}{x^2} = 0 \]
Solving for \(x\):
\[ \frac{32}{x^2} = 2 \implies x^2 = 16 \implies x = 4 \]
However, since \(x = \sin^2\theta\) must be between 0 and 1, we discard \(x = 4\) as it is not in the valid range.

### Step 5: Check the endpoints
Since the critical point is not valid, we check the endpoints of the interval \(0 < x \leq 1\).

1. As \(x \to 0^+\):
\[ y \to 2(0) + \frac{32}{0} \to \infty \]

2. At \(x = 1\):
\[ y = 2(1) + \frac{32}{1} = 2 + 32 = 34 \]

### Step 6: Use AM-GM inequality
To find the minimum value, we can also use the Arithmetic Mean-Geometric Mean (AM-GM) inequality:
\[ \frac{2\sin^2\theta + 32\csc^2\theta}{2} \geq \sqrt{2\sin^2\theta \cdot 32\csc^2\theta} \]
This simplifies to:
\[ \frac{2\sin^2\theta + 32\csc^2\theta}{2} \geq \sqrt{64} = 8 \]
Thus:
\[ 2\sin^2\theta + 32\csc^2\theta \geq 16 \]
The minimum value of \(y\) is \(16\), which occurs when \(2\sin^2\theta = 32\csc^2\theta\).

### Conclusion
The minimum value of \(2\sin^2\theta + 32\csc^2\theta\) is:
\[ \boxed{16} \]