I have seen a few solutions but I can't apply them to this particular question. The question is:
Prove that $\cot 142.5° = \sqrt{2} + \sqrt{3} - 2 - \sqrt{6}$
Help would be really appreciated.
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See https://math.stackexchange.com/questions/472594/prove-that-cot7-frac12-circ-sqrt2-sqrt3-sqrt4-sqrt6 – lab bhattacharjee Jan 13 '18 at 17:41
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I have seen this perticular solution... But I was wondering how to apply that in this case – Le Connoisseur Jan 13 '18 at 17:49
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@LeConnoisseur: so you are basically asking what is the relation between $\cot 7.5^\circ$ and $\cot (150-7.5)^\circ$, which should be simple to grasp. – Jack D'Aurizio Jan 13 '18 at 17:51
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1So, $$\cot142.5^\circ=\dfrac{1+\cos285^\circ}{\sin285^\circ}$$ $$285=360-75,75=45+30$$ – lab bhattacharjee Jan 13 '18 at 17:51
1 Answers
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use $$\cot(2x)=\frac{1}{2}\frac{\cot(x)-1}{\cot(x)}$$ and $$\cot(285^{\circ})=\sqrt{3}-2$$
Dr. Sonnhard Graubner
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