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!