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# angle multiplication and division formulae for tangent

From the angle addition formula for the tangent, we may derive formulae for tangents of multiples of angles:

$\displaystyle\tan(2x)$ | $\displaystyle={2\tan x\over 1-\tan^{2}x}$ | ||

$\displaystyle\tan(3x)$ | $\displaystyle={3\tan x-\tan^{3}x\over 1-3\tan^{2}x}$ | ||

$\displaystyle\tan(4x)$ | $\displaystyle={4\tan x-3\tan^{3}x\over 1-6\tan^{2}x+\tan^{4}x}$ |

These formulae may be derived from a recursion. Write $\tan x=w$ and write $\tan(nx)=u_{n}/v_{n}$ where the $u$’s and the $v$’s are polynomials in $w$. Then we have the initial values $u_{1}=w$ and $v_{1}=1$ and the recursions

$\displaystyle u_{{n+1}}$ | $\displaystyle=u_{n}+wv_{n}$ | ||

$\displaystyle v_{{n+1}}$ | $\displaystyle=v_{n}-wu_{n},$ |

which follow from the addition formula. Moreover, if we know the tangent of an angle and are interested in finding the tangent of a multiple of that angle, we may use our recursions directly without first having to derive the multiple angle formulae. From these recursions, one may show that the $u$’s will only involve odd powers of $w$ and the $v$’s will only involve even powers of $w$.

Proceeding in the opposite direction, one may consider bisecting an angle. Solving for $\tan x$ in the duplication formula above, one arrives at the following half-angle formula:

$\tan\left({x\over 2}\right)=\sqrt{1+{1\over\tan^{2}x}}-{1\over\tan x}$ |

Expressing the tangent in terms of sines and cosines and simplifying, one finds the following equivalent formulae:

$\tan\left({x\over 2}\right)={1-\cos x\over\sin x}={\sin x\over 1+\cos x}=\pm% \sqrt{1-\cos x\over 1+\cos x}$ |

## Mathematics Subject Classification

26A09*no label found*

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