cot-142-5-

Question Number 132880 by faysal last updated on 17/Feb/21

c o t (142.5 °) = ?

Commented by bramlexs22 last updated on 17/Feb/21

285 ° = 2 \times 142.5

285 ° = 270 ° + 15 °

\cot 285 ° = \cot 285 °

\cot (270 ° + 15 °) = \cot (2 \times 142.5 °)

- \tan 15 ° = \cot 2 α; α = 142.5 °

\Leftrightarrow \cot 2 α = - (\frac{1 - \frac{1}{\sqrt{3}}}{1 + \frac{1}{\sqrt{3}}}) = - \frac{\sqrt{3} - 1}{\sqrt{3} + 1}

\Leftrightarrow \tan 2 α = \frac{\sqrt{3} + 1}{1 - \sqrt{3}} = \frac{4 + 2 \sqrt{3}}{- 2}

\frac{2 \tan α}{1 - \tan^{2} α} = - \sqrt{3} - 2

\Rightarrow 2 \tan α = (2 + \sqrt{3}) \tan^{2} α - (\sqrt{3} + 2)

(2 + \sqrt{3}) \tan^{2} α - 2 \tan α - (\sqrt{3} + 2) = 0

\tan α = \frac{2 - \sqrt{4 + 4 {(2 + \sqrt{3})}^{2}}}{2 (2 + \sqrt{3})}

\tan α = \frac{2 - 2 \sqrt{8 + 4 \sqrt{3}}}{2 (2 + \sqrt{3})} = \frac{1 - 2 \sqrt{2 + \sqrt{3}}}{2 + \sqrt{3}}

\cot α = \frac{2 + \sqrt{3}}{1 - 2 \sqrt{2 + \sqrt{3}}} .

so \cot 142.5 ° = \frac{2 + \sqrt{3}}{1 - 2 \sqrt{2 + \sqrt{3}}} \approx - 1.303225

Commented by bramlexs22 last updated on 17/Feb/21