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Math Help - Compound angle formula help

  1. #1
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    Compound angle formula help

    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.
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  2. #2
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    Quote Originally Posted by greghunter View Post
    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)
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  3. #3
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    thanks very much
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