Sum of Arctangents/Proof
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Theorem
Let $\arctan a + \arctan b \in \openint {-\dfrac \pi 2} {\dfrac \pi 2}$
Then:
- $\arctan a + \arctan b = \map \arctan {\dfrac {a + b} {1 - a b} }$
where $\arctan$ denotes the arctangent.
Proof
Let $x = \arctan a$ and $y = \arctan b$.
Then:
\(\text {(1)}: \quad\) | \(\ds \tan x\) | \(=\) | \(\ds a\) | |||||||||||
\(\text {(2)}: \quad\) | \(\ds \tan y\) | \(=\) | \(\ds b\) | |||||||||||
\(\ds \map \tan {\arctan a + \arctan b}\) | \(=\) | \(\ds \map \tan {x + y}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac {\tan x + \tan y} {1 - \tan x \tan y}\) | Tangent of Sum | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {a + b} {1 - a b}\) | by $(1)$ and $(2)$ | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \arctan a + \arctan b\) | \(=\) | \(\ds \map \arctan {\frac {a + b} {1 - a b} }\) |
$\blacksquare$
Sources
- 1953: L. Harwood Clarke: A Note Book in Pure Mathematics ... (next): $\text V$. Trigonometry: Inverse ratios