# Simplifying $\arctan\left(\frac{\sin(\phi)+\sqrt{\frac{1}{3}}\cos(\phi)}{\cos(\phi)-\sqrt{\frac{1}{3}}\sin(\phi)}\right)$ [closed]

I'm trying to simplify the following expression involving trigonometric functions:

$$\arctan\left(\frac{\sin(\phi)+\sqrt{\frac{1}{3}}\cos(\phi)}{\cos(\phi)-\sqrt{\frac{1}{3}}\sin(\phi)}\right)$$

I'm wondering if there are any trigonometric identities or simplifications I can apply to make this expression more manageable. Any insights or suggestions on how to simplify this would be greatly appreciated. Thanks!

• Your question can be reduced to $$\arctan\left(\tan\left(\phi+\frac{\pi}{6}\right)\right)$$ Oct 11, 2023 at 7:09
• Such Questions are substantially improved by including the context of the problem. Share where the expression you want to simplify comes from, why simplifying it is important or challenging for you, or what research you did before posting here. Oct 21, 2023 at 22:19

$$\textbf{Hint:}$$ Multiply top and bottom by $$\frac{\sqrt{3}}{2}$$
$$\frac{\sin\phi \frac{\sqrt{3}}{2} + \frac{1}{2}\cos\phi}{\frac{\sqrt{3}}{2}\cos\phi - \frac{1}{2}\sin\phi}$$
and $$\cos^{-1}\frac{\sqrt{3}}{2} = \sin^{-1}\frac{1}{2} = \frac{\pi}{6}$$. Can you take it from here? Be very careful with periodicity.
First, multiply the numerator and denominator of the fraction by $$\frac12\surd3$$ to get $$\arctan\frac{\tfrac12\surd3\sin\phi+\tfrac12\cos\phi}{\tfrac12\surd3\cos\phi-\tfrac12\sin\phi}=\arctan\frac{\sin(\phi+\frac16\pi)}{\cos(\phi+\frac16\pi)}=\phi+\tfrac16\pi,$$ subject to $$-\frac23\pi<\phi<\frac13\pi$$. For $$\phi$$ outside this range, the result has to be adjusted by a suitable multiple of $$\pi$$ to bring it within the range.