If-tan-11-x-then-tan-1-

Question Number 189482 by MATHEMATICSAM last updated on 17/Mar/23

If \tan 11 ° = x then \tan 1 ° = ?

Answered by Frix last updated on 17/Mar/23

\tan 11 ° = \tan (9 ° + 2 °) = x

= \frac{\tan 9 ° + \tan 2 °}{1 - \tan 9 ° \tan 2 °} = x

\tan 2 ° = \frac{x - \tan 9 °}{1 + x \tan 9 °} = y

\tan (2 \times 1 °) = \frac{2 \tan 1 °}{1 - \tan^{2} 1 °} = y

y > 0 \land \tan 1 ° > 0 \Rightarrow

\tan 1 ° = \frac{\sqrt{y^{2} + 1} - 1}{y} =

= \frac{\sqrt{(1 + \tan^{2} 9 °) (x^{2} + 1)} - (1 + x \tan 9 °)}{x - \tan 9 °}

\tan 18 ° = \frac{\sqrt{5 (5 - 2 \sqrt{5})}}{5}

\tan 9 ° = \tan \frac{18 °}{2} = \frac{\sqrt{1 + \tan^{2} 18 °} - 1}{\tan 18 °} = \dots

= \sqrt{2 (3 + \sqrt{5})} - \sqrt{5 + 2 \sqrt{5}}

Result :

\tan 1 ° = \frac{\sqrt{(1 + \tan^{2} 9 °) (x^{2} + 1)} - (1 + x \tan 9 °)}{x - \tan 9 °}

with \tan 9 ° = \sqrt{2 (3 + \sqrt{5})} - \sqrt{5 + 2 \sqrt{5}}

Answered by mr W last updated on 17/Mar/23

\tan 11 ° = x

\tan 22 ° = \frac{2 x}{1 - x^{2}}

\tan 44 ° = \frac{2 (\frac{2 x}{1 - x^{2}})}{1 - {(\frac{2 x}{1 - x^{2}})}^{2}} = \frac{4 x (1 - x^{2})}{1 - 6 x^{2} + x^{4}}

\tan 1 ° = \tan (45 ° - 44 °) = \frac{1 - \frac{4 x (1 - x^{2})}{1 - 6 x^{2} + x^{4}}}{1 + \frac{4 x (1 - x^{2})}{1 - 6 x^{2} + x^{4}}}

= \frac{1 - 4 x - 6 x^{2} + 4 x^{3} + x^{4}}{1 + 4 x - 6 x^{2} - 4 x^{3} + x^{4}} ✓

Commented by Frix last updated on 17/Mar/23

Nice, I {didn}^{'} t think of this!

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