Question

The value of $[\sin(x) + \cos(x)]^2$ is:

• A. $1$
• B. $\sin^2(x) - \cos^2(x)$
• C. $1 + 2\sin(x)\cos(x)$
• D. $1 - 2\sin(x)\cos(x)$

Answer 1

Answer:

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

The correct answer is Option C: $1 + 2\sin(x)\cos(x)$.

To solve this, you need to combine basic algebra with a fundamental rule of trigonometry.

Step-by-Step Solution

1. Expand the expression:
Use the algebraic identity $(a + b)^2 = a^2 + 2ab + b^2$. Let $a = \sin(x)$ and $b = \cos(x)$.
$[\sin(x) + \cos(x)]^2 = \sin^2(x) + 2\sin(x)\cos(x) + \cos^2(x)$

2. Rearrange the terms:
Group the squared terms together:
$(\sin^2(x) + \cos^2(x)) + 2\sin(x)\cos(x)$

3. Apply the Pythagorean Identity:
One of the most important rules in trigonometry is that $\sin^2(x) + \cos^2(x) = 1$.

Replace that entire group with 1:

$\mathbf{1} + 2\sin(x)\cos(x)$