Question

If sin x + sin y = √3 (cos y - cos x )
Find value of sin 3x + sin 3y

Answer 1

1

Answer on Question #78863 - Math - Trigonometry

Question.

If $\sin x + \sin y = \sqrt{3} (\cos y - \cos x)$

Find value of $\sin 3x + \sin 3y$

Solution.

$$\sin x + \sin y = \sqrt{3} (\cos y - \cos x)$$ $$\frac{1}{2} \sin x + \frac{1}{2} \sin y = \frac{\sqrt{3}}{2} \cos y - \frac{\sqrt{3}}{2} \cos x$$ $$\frac{1}{2} \sin x + \frac{\sqrt{3}}{2} \cos x = \frac{\sqrt{3}}{2} \cos y - \frac{1}{2} \sin y$$ $$\frac{1}{2} = \cos \frac{\pi}{3} \quad \frac{\sqrt{3}}{2} = \sin \frac{\pi}{3}$$ $$\cos \frac{\pi}{3} \sin x + \sin \frac{\pi}{3} \cos x = \sin \frac{\pi}{3} \cos y - \cos \frac{\pi}{3} \sin y$$ $$\sin \left(\frac{\pi}{3} + x\right) = \sin \left(\frac{\pi}{3} - y\right)$$ $$\sin \left(\frac{\pi}{3} + x\right) - \sin \left(\frac{\pi}{3} - y\right) = 0$$ $$2 \sin \left(\frac{x + y}{2}\right) \cos \left(\frac{\pi}{3} + \frac{x - y}{2}\right) = 0 \Rightarrow \left( \sin \left(\frac{x + y}{2}\right) = 0 \lor \cos \left(\frac{\pi}{3} + \frac{x - y}{2}\right) = 0 \right)$$

1) $\sin \left(\frac{x + y}{2}\right) = 0 \Leftrightarrow x + y = 2n\pi, n \in \mathbb{Z} \Leftrightarrow x = -y + 2n\pi, n \in \mathbb{Z}$

$$\sin 3x + \sin 3y = \sin 3(-y + 2n\pi) + \sin 3y = \sin(-3y + 6n\pi) + \sin 3y = -\sin 3y + \sin 3y = 0, n \in \mathbb{Z}$$

2) $\cos \left(\frac{\pi}{3} + \frac{x - y}{2}\right) = 0 \Leftrightarrow x - y + \frac{2\pi}{3} = \pi + 2n\pi, n \in \mathbb{Z} \Leftrightarrow x = y + \frac{\pi}{3} + 2n\pi, n \in \mathbb{Z}$

$$\sin 3x + \sin 3y = \sin 3(y + \frac{\pi}{3} + 2n\pi) + \sin 3y = \sin(3y + \pi) + \sin 3y = -\sin 3y + \sin 3y = 0, n \in \mathbb{Z}$$

**Answer:** $\sin 3x + \sin 3y = 0$

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