Tangent of Sum of Series of Angles

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Theorem

Let $\theta_1, \theta_2, \ldots, \theta_n$ be angles.

For all $k \in \set {1, 2, 3, \ldots, n}$, let $s_k$ be defined as the sum of the product of $\map \tan {\theta_1}, \map \tan {\theta_2}, \ldots, \map \tan {\theta_n}$ taken $k$ at a time:

\(\ds s_1\) \(=\) \(\ds \sum_{i \mathop = 1}^n \map \tan {\theta_i}\)
\(\ds s_2\) \(=\) \(\ds \sum_{i \mathop = 1}^{n - 1} \sum_{j \mathop = i + 1}^n \map \tan {\theta_i} \map \tan {\theta_j}\)
\(\ds s_3\) \(=\) \(\ds \sum_{i \mathop = 1}^{n - 2} \sum_{j \mathop = i + 1}^{n - 1} \sum_{k \mathop = j + 1}^n \map \tan {\theta_i} \map \tan {\theta_j} \map \tan {\theta_k}\)
\(\ds \) \(\vdots\) \(\ds \)
\(\ds s_k\) \(=\) \(\ds \sum_{\substack {S \mathop \in \set {1, 2, \ldots, n} \\ \size S \mathop = k} } \paren {\prod_{j \mathop \in S} \tan \theta_j}\)


Then:

$\map \tan {\theta_1 + \theta_2 + \theta_3 + \cdots + \theta_n} = \dfrac {s_1 - s_3 + s_5 - \cdots} {1 - s_2 + s_4 - \cdots}$


Proof 1

First we note:

\(\ds \cos \sum_j \theta_j + i \sin \sum_j \theta_j\) \(=\) \(\ds \prod_j \paren {\cos \theta_j + i \sin \theta_j}\) Product of Complex Numbers in Polar Form
\(\ds \) \(=\) \(\ds \prod_j \cos \theta_j \prod_j \paren {1 + i \tan \theta_j}\)
\(\ds \) \(=\) \(\ds \prod_j \cos \theta_j \paren {1 + i \tan \theta_1} \times \paren {1 + i \tan \theta_2} \times \cdots \times \paren {1 + i \tan \theta_n}\) expanding the product
\(\ds \) \(=\) \(\ds \prod_j \cos \theta_j \leftparen {1 + i \paren {\tan \theta_1 + \tan \theta_2 + \cdots + \tan \theta_n} }\) multiplying out
\(\ds \) \(\) \(\, \ds + \, \) \(\ds i^2 \paren {\tan \theta_1 \tan \theta_2 + \tan \theta_1 \tan \theta_3 + \cdots + \tan \theta_{n - 1} \tan \theta_n}\)
\(\ds \) \(\) \(\, \ds + \, \) \(\ds \cdots\)
\(\ds \) \(\) \(\, \ds + \, \) \(\ds \rightparen {i^n \tan \theta_1 \tan \theta_2 \ldots \tan \theta_n}\)
\(\ds \) \(=\) \(\ds \prod_j \cos \theta_j \sum_{k \mathop = 0}^n i^k s_k\) $s_0 = 1$
\(\ds \) \(=\) \(\ds \prod_j \cos \theta_j \paren {1 - s_2 + s_4 - s_6 + \cdots} + \prod_j \cos \theta_j \paren {i s_1 - i s_3 + i s_5 - \cdots }\)
\(\text {(1)}: \quad\) \(\ds \leadsto \ \ \) \(\ds \cos \sum_j \theta_j\) \(=\) \(\ds \prod_j \cos \theta_j \paren {1 - s_2 + s_4 - s_6 + \cdots}\) equating real parts

Then:

\(\ds \cos \sum_j \theta_j + i \sin \sum_j \theta_j\) \(=\) \(\ds \prod_j \paren {\cos \theta_j + i \sin \theta_j}\) Product of Complex Numbers in Polar Form
\(\ds \leadsto \ \ \) \(\ds 1 + i \tan \sum_j \theta_j\) \(=\) \(\ds \frac {\prod_j \cos \theta_j} {\cos \sum_j \theta_j} \prod_j \paren {1 + i \tan \theta_j}\) dividing both sides by $\cos \sum_j \theta_j$
\(\ds \) \(=\) \(\ds \frac {\prod_j \cos \theta_j} {\cos \sum_j \theta_j} \paren {1 + i \tan \theta_1} \times \paren {1 + i \tan \theta_2} \times \cdots \times \paren {1 + i \tan \theta_n}\) expanding the product
\(\ds \) \(=\) \(\ds \prod_j \cos \theta_j \leftparen {1 + i \paren {\tan \theta_1 + \tan \theta_2 + \cdots + \tan \theta_n} }\) multiplying out
\(\ds \) \(\) \(\, \ds + \, \) \(\ds i^2 \paren {\tan \theta_1 \tan \theta_2 + \tan \theta_1 \tan \theta_3 + \cdots + \tan \theta_{n - 1} \tan \theta_n}\)
\(\ds \) \(\) \(\, \ds + \, \) \(\ds \cdots\)
\(\ds \) \(\) \(\, \ds + \, \) \(\ds \rightparen {i^n \tan \theta_1 \tan \theta_2 \ldots \tan \theta_n}\)
\(\ds \) \(=\) \(\ds \prod_j \cos \theta_j \sum_{k \mathop = 0}^n i^k s_k\) $s_0 = 1$
\(\ds \) \(=\) \(\ds \frac {\prod_j \cos \theta_j} {\cos \sum_j \theta_j} \paren {i s_1 - i s_3 + i s_5 - \cdots } + \frac {\prod_j \cos \theta_j} {\cos \sum_j \theta_j} \paren {1 - s_2 + s_4 - s_6 + \cdots}\)
\(\ds \leadsto \ \ \) \(\ds \frac {1 + i \tan \sum_j \theta_j} {\frac {\prod_j \cos \theta_j} {\cos \sum_j \theta_j} \paren {1 - s_2 + s_4 - s_6 + \cdots} }\) \(=\) \(\ds \frac {i \paren {s_1 - s_3 + s_5 - \cdots} } {1 - s_2 + s_4 - \cdots} + 1\) dividing both sides by $\frac {\prod_j \cos \theta_j} {\cos \sum_j \theta_j} \paren {1 - s_2 + s_4 - s_6 + \cdots}$
\(\ds \leadsto \ \ \) \(\ds \frac {\cos \sum_j \theta_j \paren {1 + i \tan \sum_j \theta_j} } {\prod_j \cos \theta_j \paren {1 - s_2 + s_4 - s_6 + \cdots} }\) \(=\) \(\ds \frac {i \paren {s_1 - s_3 + s_5 - \cdots} } {1 - s_2 + s_4 - \cdots} + 1\)
\(\ds \leadsto \ \ \) \(\ds 1 + i \tan \sum_j \theta_j\) \(=\) \(\ds \frac {i \paren {s_1 - s_3 + s_5 - \cdots} } {1 - s_2 + s_4 - \cdots} + 1\) canceling top and bottom per line $1$
\(\ds \leadsto \ \ \) \(\ds \tan \sum_j \theta_j\) \(=\) \(\ds \frac {\paren {s_1 - s_3 + s_5 - \cdots} } {1 - s_2 + s_4 - \cdots}\)

$\blacksquare$


Proof 2

First we note:

\(\ds \cos \sum_j \theta_j + i \sin \sum_j \theta_j\) \(=\) \(\ds \prod_j \paren {\cos \theta_j + i \sin \theta_j}\) Product of Complex Numbers in Polar Form
\(\ds \) \(=\) \(\ds \prod_j \cos \theta_j \prod_j \paren {1 + i \tan \theta_j}\)
\(\ds \) \(=\) \(\ds \prod_j \cos \theta_j \paren {1 + i \tan \theta_1} \times \paren {1 + i \tan \theta_2} \times \cdots \times \paren {1 + i \tan \theta_n}\) expanding the product
\(\ds \) \(=\) \(\ds \prod_j \cos \theta_j \leftparen {1 + i \paren {\tan \theta_1 + \tan \theta_2 + \cdots + \tan \theta_n} }\) multiplying out
\(\ds \) \(\) \(\, \ds + \, \) \(\ds i^2 \paren {\tan \theta_1 \tan \theta_2 + \tan \theta_1 \tan \theta_3 + \cdots + \tan \theta_{n - 1} \tan \theta_n}\)
\(\ds \) \(\) \(\, \ds + \, \) \(\ds \cdots\)
\(\ds \) \(\) \(\, \ds + \, \) \(\ds \rightparen {i^n \tan \theta_1 \tan \theta_2 \ldots \tan \theta_n}\)
\(\ds \) \(=\) \(\ds \prod_j \cos \theta_j \sum_{k \mathop = 0}^n i^k s_k\) $s_0 = 1$
\(\ds \) \(=\) \(\ds \prod_j \cos \theta_j \paren {1 - s_2 + s_4 - s_6 + \cdots} + \prod_j \cos \theta_j \paren {i s_1 - i s_3 + i s_5 - \cdots}\)
\(\text {(1)}: \quad\) \(\ds \leadsto \ \ \) \(\ds \cos \sum_j \theta_j\) \(=\) \(\ds \prod_j \cos \theta_j \paren {1 - s_2 + s_4 - s_6 + \cdots}\) equating real parts
\(\text {(2)}: \quad\) \(\ds \leadsto \ \ \) \(\ds \sin \sum_j \theta_j\) \(=\) \(\ds \prod_j \cos \theta_j \paren {s_1 - s_3 + s_5 - \cdots }\) equating imaginary parts
\(\ds \leadsto \ \ \) \(\ds \tan \sum_j \theta_j\) \(=\) \(\ds \frac {\paren {s_1 - s_3 + s_5 - \cdots} } {1 - s_2 + s_4 - \cdots}\) $\dfrac {\paren 2} {\paren 1}$

$\blacksquare$


Sources

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