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Question Number 132880 by deleteduser12 last updated on 17/Feb/21

$$cot(142.5°)=?$$

Commented by bramlexs22 last updated on 17/Feb/21

$$285°=2\times 142.5$$$$285°=270°+15°$$$$\cot 285°=\cot 285°$$$$\cot (270°+15°)=\cot (2\times 142.5°)$$$$-\tan 15°=\cot 2\alpha ;\alpha =142.5°$$$$\iff \cot 2\alpha =-\left(\frac{1-\frac{1}{\sqrt{3}}}{1+\frac{1}{\sqrt{3}}}\right)=-\frac{\sqrt{3}-1}{\sqrt{3}+1}$$$$\iff \tan 2\alpha =\frac{\sqrt{3}+1}{1-\sqrt{3}}=\frac{4+2\sqrt{3}}{-2}$$$$\frac{2\tan \alpha }{1-\tan {}^2\alpha }=-\sqrt{3}-2$$$$\Rightarrow 2\tan \alpha =(2+\sqrt{3})\tan {}^2\alpha -(\sqrt{3}+2)$$$$(2+\sqrt{3})\tan {}^2\alpha -2\tan \alpha -(\sqrt{3}+2)=0$$$$\tan \alpha =\frac{2-\sqrt{4+4{(2+\sqrt{3})}^2}}{2(2+\sqrt{3})}$$$$\tan \alpha =\frac{2-2\sqrt{8+4\sqrt{3}}}{2(2+\sqrt{3})}=\frac{1-2\sqrt{2+\sqrt{3}}}{2+\sqrt{3}}$$$$\cot \alpha =\frac{2+\sqrt{3}}{1-2\sqrt{2+\sqrt{3}}}.$$$$\mathrm{so}\cot 142.5°=\frac{2+\sqrt{3}}{1-2\sqrt{2+\sqrt{3}}}\approx -1.303225$$$$$$

Commented by bramlexs22 last updated on 17/Feb/21

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