What are some good methods for finding value of $\arctan (x- \sqrt{y}).$

Question

Don't know how to find value of $\arctan (-2-\sqrt3$)? Please give some useful insights that in 'general' (which I mentioned above in variable form) how can we find it?

Answer 1

In general, you don't get "nice" values for expressions like $\arctan(x - \sqrt y)$ or $\arctan(x + \sqrt y).$

In some cases you will get an angle that is a rational multiple of $\pi.$ For example, \begin{align} \arctan(1) &= \frac14 \pi,\\ \arctan(\sqrt3) &= \frac13 \pi. \end{align}

This is the exception rather than the rule. See $\arctan$ of a square root as a rational multiple of $\pi$ for a partial explanation.

Of course you can always construct the angle geometrically: if $x$ and $y$ are integers, classical geometry enables you to construct a line segment whose length is $x - \sqrt y$ times the length of a given line segment, and to position the new segment so the two segments meet at endpoints and are perpendicular; then you join the other endpoints with a third segment and voilá, there is an angle of measure $\arctan(x - \sqrt y).$ But that method does not necessarily lead to a good answer concerning what fraction of a full circle that angle represents.

For "nice" values of the arc tangent, you can use various techniques that basically amount to pattern recognition and guesswork. Since you're essentially guessing anyway, you can try "nice" values of the tangent function and work backward from there. The arc tangent examples above are based on the facts that $\tan\left(\frac14\pi\right)=1$ and $\tan\left(\frac13\pi\right)=1.$

Even in this direction there is only a limited amount you can do: $\tan\left(\frac17\pi\right)$ and $\tan\left(\frac19\pi\right)$ are not constructible by classical methods, so they certainly cannot be written $x - \sqrt y$ or $x + \sqrt y$ for any integers $x$ or $y.$ Constructible tangents of "nice" angles include $\tan\left(\frac15\pi\right) = \sqrt{5 - 2\sqrt 5},$ which is not quite the kind of number you want. Using the half-angle formula and the sum and difference formulas for tangents, as well as the fact that $\tan\left(\frac\pi2 - x\right) = 1/\tan(x),$ it is possible to write tangents of many other rational multiples of $\pi$ as expressions that involve addition, subtraction, and square roots starting with integers, but not so many of those will simplify to the form $x \pm \sqrt y.$

We just happen to get lucky in the case of $-2 - \sqrt 3,$ because \begin{align} \tan\left(-\frac{5}{12}\pi\right) = -\tan\left(\frac{5}{12}\pi\right) &= -\tan\left(\frac14\pi+\frac16\pi\right) \\ &= -\left(\frac{\tan\left(\frac14\pi\right) + \tan\left(\frac16\pi\right)} {1 - \tan\left(\frac14\pi\right)\tan\left(\frac16\pi\right)}\right)\\ &= -\left(\frac{1 + \frac1{\sqrt3}}{1 - (1)\left(\frac1{\sqrt3}\right)}\right)\\ &= -2 - \sqrt 3. \end{align}

I skipped several steps of the algebra that you would need to do to show the last equation in this chain of equations, but that at least is something you can solve without guessing.

You can also see similar approaches in How to calculate $\arctan(\sqrt3-2)$ to be $-15°$ by hand? (for $\arctan(-2 + \sqrt3)$ rather than $\arctan(-2 - \sqrt3),$ but these two things are closely related). Another related question is How to find precise value in terms of $ \pi $.

Answer 2

There is! In complex analysis there is the following expression for the arctangent: $$\arctan(x)=\frac{1}{2}i\text{Log}\left(\frac{1-xi}{1+xi}\right).$$ Now maybe the question remains: how to compute this logarithm? Let's look at your example: $$ \text{Log}\left(\frac{1-(-2-\sqrt{3})i}{1+(-2-\sqrt{3})i}\right)= $$ $$ \text{Log}\left({1-(-2-\sqrt{3})i}\right)-\text{Log}\left({1+(-2-\sqrt{3})i}\right) $$ where $$ \text{Log}\left({1-(-2-\sqrt{3})i}\right)=\ln|1-(-2-\sqrt{3})i|+i\text{Arg}(1-(-2-\sqrt{3})i). $$ Here, $|\cdot|$ is de modulus which means $|a+bi|=(a+bi)(a-bi)=a^2+b^2$ and Arg is the angle in $[0,2\pi)$.