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I have the trigonometric expression: $$2\sin x +2\sin \left(\frac{\pi} {3} -x\right) $$ and it should simplified in: $$\sin x + \sqrt 3 \cos x$$ but I do not know what formulas to apply. Could you tell me how to simplify it?

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Using the formula for $\sin (\alpha - \beta)$ you obtain

\begin{align} 2&\sin x +2\sin \left(\frac{\pi} {3} -x\right)\\ = 2&\sin x +2\left[\sin \left(\frac{\pi} {3}\right) \cos x - \cos\left(\frac{\pi} {3}\right) \sin x\right]\\ = 2&\sin x + 2\left[{\sqrt 3 \over 2} \cos x - \frac 1 2 \sin x\right]\\[1ex] = 2&\sin x + \sqrt 3 \cos x - \sin x\\[1em] =\ \, &\color{red}{\sin x + \sqrt 3 \cos x} \end{align}

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Use that $$\sin(x-y)=\sin(x)\cos(y)-\sin(y)\cos(x)$$

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Hint: $\sin(\frac{\pi}{3}-x)=\sin \frac{\pi}{3}\cos x-\cos \frac{\pi}{3}\sin x$

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