Trigonometry

[leonsotelo]

Prove that cotg70º+4cos70º=Sqrt(3)

M.J.

[Ale90]

Prove that cotg70º+4cos70º=Sqrt(3)

M.J.

Premessa: $ \displaystyle cos70° = cos(40°+30°) = cos40°cos30°-sen40°sen30° $$ \displaystyle = \frac{\sqrt{3}}{2}\cdot cos40° - \frac{1}{2}\cdot sen40° $
$ \displaystyle sen70° = sen(40°+30°) = sen40°cos30°+sen30°cos40° $$ \displaystyle = \frac{\sqrt{3}}{2}\cdot sen40° + \frac{1}{2}\cdot cos40° $

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$ \displaystyle cotg70°+4cos70° = \frac{cos70°}{sen70°}+4cos70° = \frac{cos70°+4sen70°cos70°}{sen70°} $

$ \displaystyle = \frac{cos70°+2sen140°}{sen70°} = \frac{cos70°+2sen40°}{sen70°} $

$ \displaystyle = \frac{\frac{\sqrt{3}}{2}\cdot cos40° + \frac{3}{2}\cdot sen40°}{\frac{1}{2}\cdot cos40° + \frac{\sqrt{3}}{2}\cdot sen40°} $$ \displaystyle = \frac{\sqrt{3}\cdot (\frac{1}{2}\cdot cos40° + \frac{\sqrt{3}}{2}\cdot sen40°)}{\frac{1}{2}\cdot cos40° + \frac{\sqrt{3}}{2}\cdot sen40°} $ $ = \sqrt{3} $