sin left(frac{pi}{3}+xright)-cos left(frac{pi | vedclass

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(D) We use the expansion formulas: $\sin(A+B) = \sin A \cos B + \cos A \sin B$ and $\cos(A+B) = \cos A \cos B - \sin A \sin B$. $\sin \left(\frac{\pi}

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$\sin x$

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(D) We use the expansion formulas: $\sin(A+B) = \sin A \cos B + \cos A \sin B$ and $\cos(A+B) = \cos A \cos B - \sin A \sin B$. $\sin \left(\frac{\pi}{3}+x\right) = \sin \frac{\pi}{3} \cos x + \cos \frac{\pi}{3} \sin x = \frac{\sqrt{3}}{2} \cos x + \frac{1}{2} \sin x$. $\cos \left(\frac{\pi}{6}+x\right) = \cos \frac{\pi}{6} \cos x - \sin \frac{\pi}{6} \sin x = \frac{\sqrt{3}}{2} \cos x - \frac{1}{2} \sin x$. Subtracting the two expressions: $\left(\frac{\sqrt{3}}{2} \cos x + \frac{1}{2} \sin x\right) - \left(\frac{\sqrt{3}}{2} \cos x - \frac{1}{2} \sin x\right)$ $= \frac{\sqrt{3}}{2} \cos x + \frac{1}{2} \sin x - \frac{\sqrt{3}}{2} \cos x + \frac{1}{2} \sin x$ $= \sin x$.

$\sin \left(\frac{\pi}{3}+x\right)-\cos \left(\frac{\pi}{6}+x\right) = $

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