Correct Answer: B (2)

To find the value of $k$, we first need to find the maximum value of the expression $\sin(x) + \sin(x + 1)$ by simplifying it using a trigonometric identity.

Step 1: Simplify the Expression using the Sum-to-Product Identity

We use the trigonometric sum-to-product identity:

$$ \sin A + \sin B = 2 \sin\left(\frac{A+B}{2}\right) \cos\left(\frac{A-B}{2}\right) $$

Let $A = x+1$ and $B = x$. Substituting these into the formula:

$\sin(x+1) + \sin(x) = 2 \sin\left(\frac{(x+1)+x}{2}\right) \cos\left(\frac{(x+1)-x}{2}\right)$

$= 2 \sin\left(\frac{2x+1}{2}\right) \cos\left(\frac{1}{2}\right)$

$= 2 \sin\left(x + \frac{1}{2}\right) \cos\left(\frac{1}{2}\right)$

Step 2: Find the Maximum Value of the Simplified Expression

The simplified expression is $2 \cos(\frac{1}{2}) \sin(x + \frac{1}{2})$.

To find its maximum value, we need to consider its components:

• The term $2\cos(\frac{1}{2})$ is a constant value (since $1/2$ is a constant angle in radians).
• The term $\sin(x + \frac{1}{2})$ is a sine function, which varies as $x$ changes.

The maximum value of any sine function, $\sin(\theta)$, is 1.

Therefore, the maximum value of our expression occurs when $\sin(x + \frac{1}{2}) = 1$.

Maximum Value = $2 \cos(\frac{1}{2}) \times 1 = 2 \cos(\frac{1}{2})$

Step 3: Determine the Value of k

The problem states that the maximum value is $k \cos(\frac{1}{2})$. We can now compare this with our calculated maximum value:

$k \cos(\frac{1}{2}) = 2 \cos(\frac{1}{2})$

By comparing the two sides, we can conclude that $k = 2$.