Let's solve the given problem step by step to prove the identity:

Problem:

Prove that
$(\sin(x) + \cos(x))^2 = 1 + \sin(2x)$ by expanding the product, applying the Pythagorean Identity and the Double-Angle Formula.

Step-by-Step Solution:

Step 1: Expand the left-hand side.

$(\sin(x) + \cos(x))^2 = \sin^2(x) + 2\sin(x)\cos(x) + \cos^2(x)$

Step 2: Apply the Pythagorean Identity:

The Pythagorean Identity states: $\sin^2(x) + \cos^2(x) = 1$ Substitute $1$ for $\sin^2(x) + \cos^2(x)$: $\sin^2(x) + \cos^2(x) + 2\sin(x)\cos(x) = 1 + 2\sin(x)\cos(x)$

Step 3: Use the Double-Angle Formula:

The Double-Angle Formula states: $\sin(2x) = 2\sin(x)\cos(x)$ Substitute $\sin(2x)$ for $2\sin(x)\cos(x)$: $1 + 2\sin(x)\cos(x) = 1 + \sin(2x)$

Final Simplified Expression:

$(\sin(x) + \cos(x))^2 = 1 + \sin(2x)$

Thus, the identity is proven.

Filled Blanks in the Steps:

1. First blank: $\cos^2(x)$
2. Second blank: $2\sin(x)\cos(x)$
3. Final answer: $1 + \sin(2x)$

Would you like a detailed breakdown of any part of the solution?

Related Questions:

1. What are some real-world applications of trigonometric identities like this one?
2. How does the Pythagorean Identity help simplify other trigonometric expressions?
3. Can this identity be proven geometrically or visually?
4. What are the other double-angle formulas for cosine and tangent?
5. What happens if we cube $(\sin(x) + \cos(x))$ instead of squaring it?

Tip:

Always verify identities by working on both sides separately to avoid introducing unnecessary errors.