Prove the trigonometric expression $\sin \left( x-\dfrac{\pi }{6} \right)+\cos \left( x-\dfrac{\pi }{3} \right)=\sqrt{3}\sin x$

Hint:Here we need to apply expansion formulas of sine and cosine in L.H.S i.e $\sin \left( A-B \right)=\sin A\cos B-\cos A\sin B$ and $\cos \left( A-B \right)=\cos A\cos B+\sin A\sin B$ in equation $\sin \left( x-\dfrac{\pi }{6} \right)+\cos \left( x-\dfrac{\pi }{3} \right)$ to get required R.HS.

Complete step-by-step answer:

Consider trigonometric expression $\sin \left( x-\dfrac{\pi }{6} \right)+\cos \left( x-\dfrac{\pi }{3} \right)=\sqrt{3}\sin x...(i)$
For proving the expressions start first by considering the left or right hand side of expressions. Here we choose left hand side of the expression $(i)$
So we consider $\sin \left( x-\dfrac{\pi }{6} \right)+\cos \left( x-\dfrac{\pi }{3} \right)...(ii)$
Now using formula $\sin \left( A-B \right)=\sin A\cos B-\cos A\sin B$ we expand the first term $\sin \left( x-\dfrac{\pi }{6} \right)$ along with substitutions $A=x$ and $B=\dfrac{\pi }{6}$
This implies that $\sin \left( x-\dfrac{\pi }{6} \right)=\sin x\cos \dfrac{\pi }{6}-\cos x\sin \dfrac{\pi }{6}...(iii)$
Now by using the values of $\cos \dfrac{\pi }{6}=\dfrac{\sqrt{3}}{2}$ and $\sin \dfrac{\pi }{6}=\dfrac{1}{2}$ and substituting these values in equation $(iii)$ we get $\sin x\cos \dfrac{\pi }{6}-\cos x\sin \dfrac{\pi }{6}=\sin x\left( \dfrac{\sqrt{3}}{2} \right)-\cos x\left( \dfrac{1}{2} \right)$
Similarly we apply the formula $\cos \left( A-B \right)=\cos A\cos B+\sin A\sin B$ in the second term of the left hand side of the expression $\cos \left( x-\dfrac{\pi }{3} \right)$
Simultaneously doing substitution as $A=x$ and $B=\dfrac{\pi }{3}$ in the expression. This implies $\cos \left( x-\dfrac{\pi }{3} \right)=\cos x\cos \left( \dfrac{\pi }{3} \right)+\sin x\sin \left( \dfrac{\pi }{3} \right)...(iv)$
Now using trigonometric values of $\cos \left( \dfrac{\pi }{3} \right)=\dfrac{1}{2}$ and $\sin \left( \dfrac{\pi }{3} \right)=\dfrac{\sqrt{3}}{2}$ in $(iv)$
By substituting these values in $(iv)$ we have $\cos x\cos \left( \dfrac{\pi }{3} \right)+\sin x\sin \left( \dfrac{\pi }{3} \right)=\dfrac{1}{2}\cos x+\dfrac{\sqrt{3}}{2}\sin x$
Now consider the simplifications of the terms. This implies $\sin \left( x-\dfrac{\pi }{6} \right)=\left( \dfrac{\sqrt{3}}{2} \right)\sin x-\left( \dfrac{1}{2} \right)\cos x$ and $\cos \left( x-\dfrac{\pi }{3} \right)=\dfrac{1}{2}\cos x+\dfrac{\sqrt{3}}{2}\sin x$
Thus after substituting these values in equation$(ii)$ the equation results into simpler term such as,
$\begin{align}
  & \sin \left( x-\dfrac{\pi }{6} \right)+\cos \left( x-\dfrac{\pi }{3} \right)=\left( \dfrac{\sqrt{3}}{2} \right)\sin x-\left( \dfrac{1}{2} \right)\cos x+\left( \dfrac{1}{2} \right)\cos x+\dfrac{\sqrt{3}}{2}\sin x \\
 & \sin \left( x-\dfrac{\pi }{6} \right)+\cos \left( x-\dfrac{\pi }{3} \right)=\left( \dfrac{\sqrt{3}}{2}+\dfrac{\sqrt{3}}{2} \right)\sin x \\
 & \sin \left( x-\dfrac{\pi }{6} \right)+\cos \left( x-\dfrac{\pi }{3} \right)=\left( \dfrac{2\sqrt{3}}{2} \right)\sin x \\
\end{align}$
By cancelling 2 from numerator and denominator we have $\sin \left( x-\dfrac{\pi }{6} \right)+\cos \left( x-\dfrac{\pi }{3} \right)=\sqrt{3}\sin x$
Clearly $\sqrt{3}\operatorname{Sin}x$ is equal to the right side of expression $(i)$ , so the result is proved.
Hence $\sin \left( x-\dfrac{\pi }{6} \right)+\cos \left( x-\dfrac{\pi }{3} \right)=\sqrt{3}\sin x$ is proved.

Note: Notice the sign is negative or positive in between the angles. Accordingly, write formulas. Also, values of the trigonometric table are important here. After using correct formulas along with right trigonometric values will easily result in the right hand side. Take care while substituting trigonometric values. If a trigonometric value is substituted then it should come before a variable. For example write $\sqrt{3}\sin x$ instead of $\sin x\left( \sqrt{3} \right)$ otherwise, root 3 gets multiplied to angle x by mistake.