How to find by hands $\arctan(\sqrt 3 + 2)$ [duplicate]

Solving an example with imaginary units.

$$\theta = \arctan(\sqrt3 + 2)$$ $$\theta = 75^o = 5\pi/12$$

Looking at the result maybe it has something to do with $$45^o$$ and $$30^o$$ angles. But how to derive the result by hands, if the only thing i have is $$\sqrt3 + 2$$

• Use that $\tan^2(\theta) + 1 = \sec^2(\theta)$ and that $\cos(2\theta) = 2\cos^2(\theta) - 1.$ Jan 5, 2022 at 16:50
• Related (Probably a duplicate)
– ACB
Jan 5, 2022 at 18:16
• @ACB Arguable whether a duplicate. I distinguish between verifying an answer and deriving an answer. Jan 6, 2022 at 7:59

It's easy to find $$\angle ABC=75^\circ$$.
We will use the half-angle formula for tangent: $$\tan\frac{\theta}{2} = \frac{1-\cos(\theta)}{\sin(\theta)} .$$ We want to get $$2+\sqrt{3}$$. Remembering the basic values of sine and cosine, I see that $$2+\sqrt3 = \frac{1+\frac{\sqrt{3}}{2}}{\frac12} = \frac{1-\cos\frac{5\pi}{6}}{\sin\frac{5\pi}{6}} = \tan\frac{5\pi}{12}$$ and therefore $$\arctan(2+\sqrt3) = \frac{5\pi}{12}$$
• when you wrote $\frac{1+\frac{\sqrt3}2}{\frac1 2}$ you took $\theta = 60^o$ . But why? What leads to choosing exactly this value? Jan 5, 2022 at 22:35
• That is the one I already know with value $\sqrt{3}$ in it. It seemed likely, since $\sqrt{3}$ was already in the problem. Actually, I took $\theta = 75^\circ$ so that cosine would be negative. Jan 6, 2022 at 0:15