Step 1: Look at what we must add. We want the largest value of $\sin x + \sin(x+1)$. Adding two sines is easier with a known identity.
Step 2: Recall the sum-to-product identity. \[ \sin A + \sin B = 2 \sin\frac{A+B}{2} \cos\frac{A-B}{2} \] Step 3: Plug in $A = x$ and $B = x+1$. The average is $\frac{A+B}{2} = x + \frac12$ and the half difference is $\frac{A-B}{2} = -\frac12$. \[ \sin x + \sin(x+1) = 2 \sin\left(x + \tfrac12\right) \cos\frac12 \] Note $\cos\left(-\frac12\right) = \cos\frac12$.
Step 4: Find the biggest the sine part can be. A sine value never goes above $1$, so $\sin\left(x + \frac12\right)$ is at most $1$.
Step 5: Put in that maximum. \[ \text{Maximum} = 2 \times 1 \times \cos\frac12 = 2\cos\frac12 \] Step 6: Compare with $k\cos\frac12$. Matching $2\cos\frac12$ with $k\cos\frac12$ gives $k = 2$. \[ \boxed{k = 2} \]