4 sinθ sin $\left(\frac{\pi }{3} + \theta \right)$ sin $\left(\frac{\pi }{3} – \theta \right)$ =sin3θ
Prove that, 4 sinθ sin(π/3+θ ) sin(π/3-θ) = sin3θ
Solution:
4 sinθ sin $\left(\frac{\pi }{3} + \theta \right)$ sin$\left(\frac{\pi }{3} – \theta \right)$
=2 sinθ {2 sin$\left(\frac{\pi }{3} + \theta \right)$ sin$\left(\frac{\pi }{3} – \theta \right)$ }
=2 sinθ $\left\{\cos \left(\frac{\pi }{3} + \theta – \frac{\pi }{3} + \theta \right) – cos\left(\frac{\pi }{3} + \theta + \frac{\pi }{3} – \theta \right)\right\}$
=2 sinθ $(\cos 2\theta – cos\frac{2\pi }{3})$
=2 sinθ $\left\{\cos 2\theta – cos\left(\pi – \frac{\pi }{3}\right)\right\}$
=2 sinθ (cos2θ+cos $\frac{\pi }{3}$)
=2 sinθ (cos2 θ +$\frac{1}{2}$ )
=2sinθ cos2θ + sinθ
= sin(θ +2θ)+sin(θ-2 θ)+sin θ
= sin3θ –sinθ +sinθ
= sin3θ [Proved]