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Let's solve and verify each of these trigonometric identities, justifying each step as requested.

### 1. \((\sin \theta - \cos \theta)^2 = 1 - \sin(2\theta)\)

#### Step-by-step solution:
1. Expand the left-hand side (LHS):
   \[
   (\sin \theta - \cos \theta)^2 = \sin^2 \theta - 2 \sin \theta \cos \theta + \cos^2 \theta
   \]
2. Use the Pythagorean identity: \(\sin^2 \theta + \cos^2 \theta = 1\).
   \[
   = 1 - 2 \sin \theta \cos \theta
   \]
3. Apply the double angle identity for sine: \(\sin(2\theta) = 2 \sin \theta \cos \theta\).
   \[
   = 1 - \sin(2\theta)
   \]
Thus, the left-hand side simplifies to \(1 - \sin(2\theta)\), which matches the right-hand side (RHS). Verified.

---

### 2. \(\frac{1 + \cos(2\theta)}{\sin(2\theta)} \cdot \tan^2 \theta = \tan \theta\)

#### Step-by-step solution:
1. Recall that \(\tan^2 \theta = \frac{\sin^2 \theta}{\cos^2 \theta}\), so the expression becomes:
   \[
   \frac{1 + \cos(2\theta)}{\sin(2\theta)} \cdot \frac{\sin^2 \theta}{\cos^2 \theta}
   \]
2. Use the double angle identities: \(\cos(2\theta) = 1 - 2 \sin^2 \theta\) and \(\sin(2\theta) = 2 \sin \theta \cos \theta\).
   \[
   = \frac{1 + (1 - 2 \sin^2 \theta)}{2 \sin \theta \cos \theta} \cdot \frac{\sin^2 \theta}{\cos^2 \theta}
   \]
   Simplifying the numerator of the first fraction:
   \[
   = \frac{2 - 2 \sin^2 \theta}{2 \sin \theta \cos \theta} \cdot \frac{\sin^2 \theta}{\cos^2 \theta}
   \]
3. Factor out a 2 from the numerator:
   \[
   = \frac{2(1 - \sin^2 \theta)}{2 \sin \theta \cos \theta} \cdot \frac{\sin^2 \theta}{\cos^2 \theta}
   \]
4. Since \(1 - \sin^2 \theta = \cos^2 \theta\), substitute:
   \[
   = \frac{2 \cos^2 \theta}{2 \sin \theta \cos \theta} \cdot \frac{\sin^2 \theta}{\cos^2 \theta}
   \]
5. Cancel common terms:
   \[
   = \frac{\sin^2 \theta}{\sin \theta \cos \theta}
   \]
6. Simplify:
   \[
   = \frac{\sin \theta}{\cos \theta} = \tan \theta
   \]
Thus, both sides are equal. Verified.

---

### 3. \(\sin(3x) = 3 \sin x \cos^2 x - \sin^3 x\)

#### Step-by-step solution:
1. Use the triple angle formula for sine: \(\sin(3x) = 3 \sin x - 4 \sin^3 x\).
2. Re-arrange the formula so that it becomes:
   \[
   \sin(3x) = 3 \sin x (1 - \sin^2 x) - \sin^3 x
   \]
3. Use the Pythagorean identity: \(1 - \sin^2 x = \cos^2 x\), so we substitute:
   \[
   \sin(3x) = 3 \sin x \cos^2 x - \sin^3 x
   \]
Thus, both sides are equal. Verified.

---

### Do you have any questions, or would you like more details on any of these steps?

Here are some related questions you might want to explore:
1. How can you verify trigonometric identities more efficiently?
2. What are other double angle identities besides \(\sin(2\theta)\)?
3. How does the expansion of \((\sin \theta - \cos \theta)^2\) compare to other squared binomial expansions?
4. What are other forms of the triple angle identities for sine and cosine?
5. How are trigonometric identities used in calculus?

**Tip:** Always check if you can simplify the trigonometric identities using known formulas like Pythagorean, double-angle, or sum-to-product identities!

Verify, justifying each step.
1. (sin(θ) − cos(θ))^2 = 1 − sin(2θ)
2. (1 + cos(2θ)) / sin(2θ) * tan^2(θ) = tan(θ)
3. sin(3x) = 3 sin(x) cos^2(x) − sin^3(x)

Explore step-by-step solutions to verify trigonometric identities such as (sin(θ) − cos(θ))^2 = 1 − sin(2θ) and sin(3x) = 3 sin(x) cos^2(x) − sin^3(x), utilizing double angle, triple angle, and Pythagorean identities. Suitable for Precalculus students in Grades 11-12.

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