If `pi lt alpha lt (3pi)/2` then find the value of expression `sqrt(4 sin^4 alpha+ sin^2 2alpha)+ 4 cos^2(pi/4- alpha/2)`

Answer

To solve the expression \( \sqrt{4 \sin^4 \alpha + \sin^2 2\alpha} + 4 \cos^2\left(\frac{\pi}{4} - \frac{\alpha}{2}\right) \) given that \( \pi < \alpha < \frac{3\pi}{2} \), we will follow these steps:

### Step 1: Simplify \( \sin^2 2\alpha \)
We know that:
\[ \sin 2\alpha = 2 \sin \alpha \cos \alpha \]
Thus,
\[ \sin^2 2\alpha = (2 \sin \alpha \cos \alpha)^2 = 4 \sin^2 \alpha \cos^2 \alpha \]

### Step 2: Substitute \( \sin^2 2\alpha \) into the expression
Now substituting this into the original expression:
\[ \sqrt{4 \sin^4 \alpha + 4 \sin^2 \alpha \cos^2 \alpha} \]

### Step 3: Factor out common terms
We can factor out \( 4 \sin^2 \alpha \):
\[ \sqrt{4 \sin^2 \alpha (\sin^2 \alpha + \cos^2 \alpha)} = \sqrt{4 \sin^2 \alpha (1)} = \sqrt{4 \sin^2 \alpha} = 2 |\sin \alpha| \]

### Step 4: Determine the sign of \( \sin \alpha \)
Since \( \alpha \) is in the interval \( (\pi, \frac{3\pi}{2}) \), \( \sin \alpha \) is negative. Therefore, \( |\sin \alpha| = -\sin \alpha \).

Thus,
\[ 2 |\sin \alpha| = -2 \sin \alpha \]

### Step 5: Simplify \( 4 \cos^2\left(\frac{\pi}{4} - \frac{\alpha}{2}\right) \)
Using the cosine identity:
\[ \cos\left(\frac{\pi}{4} - \frac{\alpha}{2}\right) = \frac{\sqrt{2}}{2}(\cos\frac{\alpha}{2} + \sin\frac{\alpha}{2}) \]
Thus,
\[ \cos^2\left(\frac{\pi}{4} - \frac{\alpha}{2}\right) = \frac{1}{2}(\cos^2\frac{\alpha}{2} + \sin^2\frac{\alpha}{2} + 2\sin\frac{\alpha}{2}\cos\frac{\alpha}{2}) = \frac{1}{2}(1 + \sin \alpha) \]
So,
\[ 4 \cos^2\left(\frac{\pi}{4} - \frac{\alpha}{2}\right) = 2(1 + \sin \alpha) \]

### Step 6: Combine the results
Now, combining the results from Step 4 and Step 5:
\[ -2 \sin \alpha + 2(1 + \sin \alpha) = -2 \sin \alpha + 2 + 2 \sin \alpha = 2 \]

### Final Result
Thus, the value of the expression is:
\[ \boxed{2} \]