The maximum value of \(\sin(x) + \sin(x+1)\) is \(k \cos{\frac{1}{2}}\). Then the value of k is
Step-by-step Solution:
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.
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:
$\sin(x+1) + \sin(x) = 2 \sin\left(\frac{(x+1)+x}{2}\right) \cos\left(\frac{(x+1)-x}{2}\right)$
$= 2 \sin\left(\frac{2x+1}{2}\right) \cos\left(\frac{1}{2}\right)$
$= 2 \sin\left(x + \frac{1}{2}\right) \cos\left(\frac{1}{2}\right)$
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 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 = $2 \cos(\frac{1}{2}) \times 1 = 2 \cos(\frac{1}{2})$
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 $k = 2$.