Tangent of Sum of Three Angles
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Theorem
- $\map \tan {A + B + C} = \dfrac {\tan A + \tan B + \tan C - \tan A \tan B \tan C} {1 - \tan B \tan C - \tan C \tan A - \tan A \tan B}$
Proof 1
\(\ds \map \sin {A + B + C}\) | \(=\) | \(\ds \sin A \cos B \cos C + \cos A \sin B \cos C + \cos A \cos B \sin C - \sin A \sin B \sin C\) | Sine of Sum of Three Angles | |||||||||||
\(\ds \map \cos {A + B + C}\) | \(=\) | \(\ds \cos A \cos B \cos C - \sin A \sin B \cos C - \sin A \cos B \sin C - \cos A \sin B \sin C\) | Cosine of Sum of Three Angles | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \map \tan {A + B + C}\) | \(=\) | \(\ds \dfrac {\sin A \cos B \cos C + \cos A \sin B \cos C + \cos A \cos B \sin C - \sin A \sin B \sin C} {\cos A \cos B \cos C - \sin A \sin B \cos C - \sin A \cos B \sin C - \cos A \sin B \sin C}\) | Tangent is Sine divided by Cosine | ||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {\frac {\sin A \cos B \cos C} {\cos A \cos B \cos C} + \frac {\cos A \sin B \cos C} {\cos A \cos B \cos C} + \frac {\cos A \cos B \sin C} {\cos A \cos B \cos C} - \frac {\sin A \sin B \sin C} {\cos A \cos B \cos C} } {\frac {\cos A \cos B \cos C} {\cos A \cos B \cos C} - \frac {\sin A \sin B \cos C} {\cos A \cos B \cos C} - \frac {\sin A \cos B \sin C} {\cos A \cos B \cos C} - \frac {\cos A \sin B \sin C} {\cos A \cos B \cos C} }\) | dividing numerator and denominator by $\cos A \cos B \cos C$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {\tan A + \tan B + \tan C - \tan A \tan B \tan C} {1 - \tan B \tan C - \tan C \tan A - \tan A \tan B}\) | Tangent is Sine divided by Cosine and simplifying |
$\blacksquare$
Proof 2
\(\ds \map \tan {A + B + C}\) | \(=\) | \(\ds \dfrac {\tan A + \map \tan {B + C} } {1 - \tan A \tan {B + C} }\) | Tangent of Sum | |||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {\tan A + \frac {\tan B + \tan C} {1 - \tan B \tan C} } {1 - \tan A \frac {\tan B + \tan C} {1 - \tan B \tan C} }\) | Tangent of Sum | |||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {\tan A \paren {1 - \tan B \tan C} + \tan B + \tan C} {\paren {1 - \tan B \tan C} - \tan A \paren {\tan B + \tan C} }\) | multiplying top and bottom by $1 - \tan B \tan C$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {\tan A + \tan B + \tan C - \tan A \tan B \tan C} {1 - \tan B \tan C - \tan C \tan A - \tan A \tan B}\) | simplification |
$\blacksquare$
Proof 3
This is a special case of Tangent of Sum of Series of Angles, for $n = 3$.
$\blacksquare$
Also see
Sources
- 1953: L. Harwood Clarke: A Note Book in Pure Mathematics ... (previous) ... (next): $\text V$. Trigonometry: Formulae $(20)$