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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 $\sqrt3 + 2$

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

2 Answers 2

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It's easy to find $\angle ABC=75^\circ$.

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    $\begingroup$ appreciate. it's beautiful. $\endgroup$
    – kertal
    Jan 5, 2022 at 22:20
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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 $$\arctan(2+\sqrt3) = \frac{5\pi}{12}$$

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  • $\begingroup$ when you wrote $\frac{1+\frac{\sqrt3}2}{\frac1 2}$ you took $\theta = 60^o$ . But why? What leads to choosing exactly this value? $\endgroup$
    – kertal
    Jan 5, 2022 at 22:35
  • $\begingroup$ 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. $\endgroup$
    – GEdgar
    Jan 6, 2022 at 0:15

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