How do you show that 2 \sin x \cos x = \sin 2x? is true for (5pi)/6?

Oct 24, 2014

To prove it, evaluate the left-hand side and the right-hand side separately and show that they are the same.

(LHS)$= 2 \sin \left(\frac{5 \pi}{6}\right) \cos \left(\frac{5 \pi}{6}\right) = 2 \cdot \frac{1}{2} \cdot \left(- \frac{\sqrt{3}}{2}\right) = - \frac{\sqrt{3}}{2}$

(RHS)$= \sin \left(2 \cdot \frac{5 \pi}{6}\right) = \sin \left(\frac{5 \pi}{3}\right) = - \frac{\sqrt{3}}{2}$

Hence, (LHS)$=$(RHS) for $x = \frac{5 \pi}{6}$.

I hope that this was helpful.