IME / ITA ⇒ (ITA - 1987) Trigonometria

[leoguedes]

O valor de x>0 que satisfaz a equação [tex3]\sqrt{x}=\tan \left(\frac{\pi}{12}\right)[/tex3] é:

a) 4 [tex3]\sqrt{3}[/tex3]
b) 5-4 [tex3]\sqrt{3}[/tex3]
c) 7-[tex3]\sqrt{3}[/tex3]
d) 7-4 [tex3]\sqrt{3}[/tex3]
e) 9-4 [tex3]\sqrt{3}[/tex3]

Resposta

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d)

[csmarceloMOD]

[tex3]\sqrt{x}=\tan\left(\frac{\pi}{12}\right)\Rightarrow x=\tan^2\left(\frac{\pi}{12}\right)[/tex3]

Da fórmula do arco metade:

[tex3]\tan\left(\frac{\pi}{12}\right)=\pm\sqrt{\frac{1-\cos\left(\frac{\pi}{6}\right)}{1+\cos\left(\frac{\pi}{6}\right)}}[/tex3]

Portanto,

[tex3]\tan^2\left(\frac{\pi}{12}\right)=\frac{1-\cos\left(\frac{\pi}{6}\right)}{1+\cos\left(\frac{\pi}{6}\right)}=\frac{1-\frac{\sqrt{3}}{2}}{1+\frac{\sqrt{3}}{2}}=7-4\sqrt{3}[/tex3]