2

$$\tan(\theta_1+\theta_2+\cdots+\theta_n)=\frac{S_1-S_3+S_5+\cdots}{1-S_2+S_4+\cdots}$$

where $S_i$ denotes the sum of product of tangent of angles taken $i$ at a time.

For example,$$\tan(\theta_1+\theta_2)=\frac{S_1}{1-S_2}=\frac{\tan\theta_1+\tan\theta_2} {1-\tan\theta_1\tan\theta_2}$$

(This formula is given in my textbook with no derivation or background)

How to derive this?

Blue
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  • Please do not upvote the question for it shows no effort/research. –  Mar 04 '17 at 16:59
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    http://math.stackexchange.com/questions/346368/sum-of-tangent-functions-where-arguments-are-in-specific-arithmetic-series – lab bhattacharjee Mar 04 '17 at 17:03
  • "Please do not upvote the question for it shows no effort/research." I realize it's a fine line but I think genuine confusion and honestly having no idea how to start is legitimate, whereas laziness is not. And for a vague question such as "how to derive this" I think an answer of "By the definition of tan as sin/cos and the identities for sin(a +b) and cos(a+b)" is a valid and complete answer. – fleablood Mar 04 '17 at 17:15

2 Answers2

1

Prove that is true for $2$ angles, then consider it true for $(n-1)$ angles and prove for $n$ angles. If you prove it for $2$ angles, it will be easy to prove it for $n$ angles (knowing it is true for $(n-1)$ angles) by considering $\theta_1+...+\theta_{n-1}$ is one angle, and $\theta_n$ is the other.

Del
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Angelo Saadeh
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0

Once you derive the first the rest follow by induction.

As for the first $\tan (a + b) = \frac {\sin (a + b)}{\cos (a+b)} = \frac {\sin a\cos b + \cos a \sin b}{\cos a \cos b - \sin a \sin b}$

Divide top and bottom by $\cos a \cos b$ and you get the result.

fleablood
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