The expression 
1
+
sin
2
α
cos
(
2
α
−
2
π
)
tan
(
α
−
3
(
π
2
)
)
−
1
4
sin
2
α
(
cot
(
α
2
)
+
cot
(
3
(
π
2
)
+
α
2
)
)
 when simplified reduces to

Let's simplify the expression step by step:

\[
1 + \frac{\sin^2 \alpha}{\cos(2\alpha - 2\pi)} \tan\left(\alpha - 3\left(\frac{\pi}{2}\right)\right) - \frac{1}{4} \sin^2 \alpha \left(\cot\left(\frac{\alpha}{2}\right) + \cot\left(3\left(\frac{\pi}{2}\right) + \frac{\alpha}{2}\right)\right)
\]

### Step 1: Simplify the trigonometric arguments
#### 1.1. Simplify \( \cos(2\alpha - 2\pi) \):
We know that subtracting \(2\pi\) from any angle results in the same value because of periodicity:
\[
\cos(2\alpha - 2\pi) = \cos(2\alpha)
\]

#### 1.2. Simplify \( \tan\left(\alpha - 3\left(\frac{\pi}{2}\right)\right) \):
Shifting by \( \frac{3\pi}{2} \) corresponds to a phase shift of 270 degrees:
\[
\tan\left(\alpha - 3\left(\frac{\pi}{2}\right)\right) = \tan\left(\alpha + \frac{\pi}{2}\right)
\]
Using the identity \( \tan\left(\theta + \frac{\pi}{2}\right) = -\cot(\theta) \):
\[
\tan\left(\alpha + \frac{\pi}{2}\right) = -\cot(\alpha)
\]

So now the expression becomes:
\[
1 + \frac{\sin^2 \alpha}{\cos(2\alpha)} \cdot (-\cot(\alpha)) - \frac{1}{4} \sin^2 \alpha \left(\cot\left(\frac{\alpha}{2}\right) + \cot\left(3\left(\frac{\pi}{2}\right) + \frac{\alpha}{2}\right)\right)
\]

### Step 2: Simplify the cotangent terms
#### 2.1. Simplify \( \cot\left(3\left(\frac{\pi}{2}\right) + \frac{\alpha}{2}\right) \):
Using the same identity for a phase shift by \(3\pi/2\), we get:
\[
\cot\left(3\left(\frac{\pi}{2}\right) + \frac{\alpha}{2}\right) = -\tan\left(\frac{\alpha}{2}\right)
\]

Thus, the expression simplifies to:
\[
1 - \frac{\sin^2 \alpha \cot(\alpha)}{\cos(2\alpha)} - \frac{1}{4} \sin^2 \alpha \left(\cot\left(\frac{\alpha}{2}\right) - \tan\left(\frac{\alpha}{2}\right)\right)
\]

### Step 3: Further simplifications
1. The term \( \sin^2 \alpha \cot(\alpha) \) can be simplified using the identity \( \cot(\alpha) = \frac{\cos(\alpha)}{\sin(\alpha)} \):
   \[
   \sin^2 \alpha \cot(\alpha) = \sin\alpha \cos\alpha
   \]

   Hence, the term becomes:
   \[
   \frac{\sin\alpha \cos\alpha}{\cos(2\alpha)}
   \]

2. \( \cos(2\alpha) = 2\cos^2(\alpha) - 1 \), which can be used if needed.

At this point, a simplified form of the original expression is:
\[
1 - \frac{\sin \alpha \cos \alpha}{\cos(2\alpha)} - \frac{1}{4} \sin^2 \alpha \left(\cot\left(\frac{\alpha}{2}\right) - \tan\left(\frac{\alpha}{2}\right)\right)
\]

This expression doesn't immediately simplify further without additional substitutions or assumptions. Depending on specific values of \( \alpha \), this can be evaluated numerically.

Would you like me to continue the simplification, or do you have any specific questions about the process?

---

Here are 5 related questions that can expand on this:

1. How do trigonometric identities such as angle addition and double-angle formulas help simplify expressions?
2. What is the significance of periodicity in trigonometric functions?
3. How does the cotangent function behave with phase shifts like \( \frac{\pi}{2} \) and \( \pi \)?
4. How can numerical methods be used to solve trigonometric expressions when exact simplification is challenging?
5. How can trigonometric expressions like this be applied to solve problems in physics, particularly in wave mechanics?

**Tip**: When simplifying trigonometric expressions, always check for opportunities to apply standard identities, such as double angle or sum formulas, to make the expression more manageable.

The expression 1 + sin^2(α)cos(2α−2π)tan(α−3(π/2))−(1/4)sin^2(α)(cot(α/2)+cot(3(π/2)+α/2)) when simplified reduces to

This problem involves simplifying a complex trigonometric expression that includes sine, cosine, and tangent functions, along with angle shifts and trigonometric identities like double-angle and cotangent formulas. Suitable for college-level students, this detailed solution breaks down each step, applying key trigonometric theorems.

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