the equation $sinleft(\frac{pi}{6}-x\right)$ is equal to ___.\n\na. $\frac{1}{2}(sqrt{3}cos x - sin x)$\nb…

Explanation:

Step1: Use the sine - difference formula

The formula for $\sin(A - B)=\sin A\cos B-\cos A\sin B$. Here $A = \frac{\pi}{6}$ and $B=x$. So $\sin(\frac{\pi}{6}-x)=\sin\frac{\pi}{6}\cos x-\cos\frac{\pi}{6}\sin x$.

Step2: Substitute the values of trigonometric functions

We know that $\sin\frac{\pi}{6}=\frac{1}{2}$ and $\cos\frac{\pi}{6}=\frac{\sqrt{3}}{2}$. Then $\sin\frac{\pi}{6}\cos x-\cos\frac{\pi}{6}\sin x=\frac{1}{2}\cos x-\frac{\sqrt{3}}{2}\sin x=\frac{1}{2}(\cos x - \sqrt{3}\sin x)$.

Answer:

B. $\frac{1}{2}(\cos x-\sqrt{3}\sin x)$