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I had a problem to express $\tan kx$ as a function of $\tan x$. For example, $\tan 3x=(3 \tan x−\tan^ 3x)/(1−3\tan^2x)$. But in general case, how can I express for example $\tan 10x$ as a powers of $\tan x$?

I saw a Chebyshev method from http://www.trans4mind.com/personal_development/mathematics/trigonometry/multipleAnglesRecursiveFormula.htm but how do this works in practice?

curious
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2 Answers2

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$kx$ is the argument of $(1+i \tan x)^k$, so to get $\tan kx$ you can work out the real and imaginary parts of that using the binomial theorem and divide:

$$ \tan kx = \frac{\sum_{n=0}^{\lceil k/2 \rceil-1} \binom{k}{2n+1}(-1)^n \tan^{2n+1}x }{\sum_{n=0}^{\lfloor k/2\rfloor} \binom{k}{2n}(-1)^n\tan^{2n}x} $$

hmakholm left over Monica
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    IOW, just get the binomial coefficients for $k$, and alternate them from the denominator to the numerator, starting at the denominator, and alternate the signs within the denominator & numerator, with both starting on positive. Eg for $k=10$, writing $t$ for $\tan(x)$, we get $\tan(10x)=\frac{10t^1-120t^3+252t^5-120t^7+10t^9}{1t^0-45t^2+210t^4-210t^6+ 45 t^8 - 1t^{10}}$. – PM 2Ring Sep 11 '15 at 13:46
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There is also the recurrence relation: $$ \tan\,((n{+}1)\theta) = \frac{\tan (n\theta) + \tan \theta}{1 - \tan (n\theta)\,\tan \theta} $$ which implies that $\tan n \theta$ is a rational function of $\tan \theta$ with integer coefficients.

lhf
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