Research website of Vyacheslav Gorchilin
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Formula. Math. Trigonometry
The sum and difference of inverse functions
$\arcsin x + \arcsin y$
$\arccos x + \arccos y$
$\arctan x + \arctan y$
$\arctan x + \arctan y$
$\arctan x - \arctan y$
$\arctan x$
$\arccos x$
$\arctan \frac12 + \arctan \frac13$
The amount of arctinus$\arcsin x + \arcsin y = \begin{cases} T,\quad xy \le 0 \\ \pi-T, x \ge 0, y \ge 0 \\ -\pi-T, x \lt 0, y \lt 0 \end{cases} \quad T = \arcsin\left(x\sqrt{1-y^2} + y\sqrt{1-x^2}\right)$
The amount of arccosine$\arccos x + \arccos y = \begin{cases} T, \quad x+y \ge 0 \\ 2\pi-T, x+y \lt 0 \end{cases} \quad T = \arccos\left(xy-\sqrt{1-y^2}\sqrt{1-x^2}\right)$
The sum of the arc tangent$\arctan x + \arctan y = \begin{cases} T, \quad xy \gt -1 \\ \pi+T, x \gt 0, xy \lt -1 \\ -\pi+T, x \lt 0, xy \gt -1 \end{cases} \quad T = \arctan \frac{x-y}{1+xy}$
The sum of the arc tangent (2)$\arctan x + \arctan y = \arccos {1-xy \over \sqrt{1+y^2}\sqrt{1+x^2}}$
The difference between the arc tangent$\arctan x - \arctan y = \arccos {1+xy \over \sqrt{1+y^2}\sqrt{1+x^2}}$
Representation of a tangent using the inverse cosine$\arctan x = \arccos {1 \over \sqrt{1+x^2}}$
Representation of a tangent using the inverse cosine$\arccos x = \begin{cases} T, \quad 0 \le x \le 1 \\ \pi+T, \quad -1 \le x \le 0 \end{cases} \quad T = \arctan {\sqrt{1-x^2} \over x}$
$\arctan \frac12 + \arctan \frac13 = \frac{\pi}{4}$