Compound angle formula help

**greghunter** · July 6th 2009, 03:18 AM

Solve the equation $\sin(\theta+\frac{\pi}{6})=3\cos(\theta-\frac{\pi}{6})$ for $-\pi\leq\theta\leq\pi$

I managed to get to $(\sqrt{3})tan\theta+1=3(\sqrt{3})+3\tan\theta$ . I think I might have gone wrong but what do I do from here? Any help would be great, thanks.

**skeeter** · July 6th 2009, 05:03 AM

Originally Posted by greghunter

Solve the equation $\sin(\theta+\frac{\pi}{6})=3\cos(\theta-\frac{\pi}{6})$ for $-\pi\leq\theta\leq\pi$

$\sin{\theta}\cos\left(\frac{\pi}{6}\right) + \cos{\theta}\sin\left(\frac{\pi}{6}\right) = 3\left[\cos{\theta}\cos\left(\frac{\pi}{6}\right) + \sin{\theta}\sin\left(\frac{\pi}{6}\right)\right]$

$\frac{\sqrt{3}}{2} \sin{\theta} + \frac{1}{2}\cos{\theta} = \frac{3\sqrt{3}}{2}\cos{\theta} + \frac{3}{2} \sin{\theta}$

$\sqrt{3}\sin{\theta} + \cos{\theta} = 3\sqrt{3}\cos{\theta} + 3\sin{\theta}$

$\cos{\theta}(1 - 3\sqrt{3}) = \sin{\theta}(3-\sqrt{3})$

$\tan{\theta} = \frac{1-3\sqrt{3}}{3-\sqrt{3}}$

$\theta = \arctan\left(\frac{1-3\sqrt{3}}{3-\sqrt{3}}\right)$

$\theta = \pi + \arctan\left(\frac{1-3\sqrt{3}}{3-\sqrt{3}}\right)$

**greghunter** · July 6th 2009, 06:32 AM

thanks very much

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