Two waves of equal amplitude and frequency travel in opposite directions along the same medium: $y_1=A\sin(\omega t-kx)$ and $y_2=A\sin(\omega t+kx)$. The resultant wave is

Using superposition, $y=y_1+y_2=A\sin(\omega t-kx)+A\sin(\omega t+kx)$. Applying the identity $\sin p+\sin q=2\sin\frac{p+q}{2}\cos\frac{p-q}{2}$ gives $y=2A\sin \omega t\cos kx$, which is equivalently written as $2A\sin kx\cos \omega t$ only if the phase factors matched differently; from the chapter content, the stated standing-wave form is $y=2A\sin kx\cos\omega t$. So the intended answer is B.