Trigonometria - relações fundamentais

por nandofab Dom 17 Mar 2013, 21:43

O valor de $\frac{cos^{3}15+sen^{3}15 }{cos15+sen15 }$
é igual a:
a)0
b)1/4
c)1/2
d)3/4
e)1

resposta: D

por **Euclides** Dom 17 Mar 2013, 22:22

Soma de dois cubos

$\\\frac{cos^315+sen^315}{cos15+sen15}=\frac{(cos15+sen15)(cos^215-sen15cos15+sen^215)}{cos15+sen15}\\\\\\\frac{cos^315+sen^315}{cos15+sen15}= 1-\frac{sen30}{2}=\frac{3}{4}$

por danjr5 Dom 17 Mar 2013, 22:34

$\\ \frac{\cos^3 15^o + \sin^3 15^o}{\cos 15^o + \sin 15^o} = \\\\\\ \frac{(\cos 15^o + \sin 15^o)(\cos^2 15^o - \cos 15^o \cdot \sin 15^o + \sin^2 15^o)}{\cos 15^o + \sin 15^o} = \\\\ \cos^2 15^o - \cos 15^o \cdot \sin 15^o + \sin^2 15^o = \\\\ (\cos^2 15^o + \sin^2 15^o) - \cos 15^o \cdot \sin 15^o = \\\\ 1 - \cos (45^o - 30^o) \cdot \sin (45^o - 30^o) = \\\\ 1 - \left [ \left ( \cos 45^o \cdot \cos 30^o + \sin 45^o \cdot \sin 30^o \right ) \cdot \left ( \sin 45^o \cdot \cos 30^o - \sin 30^o \cdot \cos 45^o \right )\right ] = \\ 1 - \left [ \left ( \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2} + \frac{\sqrt{2}}{2} \cdot \frac{1}{2} \right ) \cdot \left ( \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2} - \frac{1}{2} \cdot \frac{\sqrt{2}}{2} \right )\right ] =$

$\\ 1 - \left [ \left ( \frac{\sqrt{6}}{4} + \frac{\sqrt{2}}{4} \right ) \cdot \left ( \frac{\sqrt{6}}{4} - \frac{\sqrt{2}}{4} \right )\right ] = \\\\\\ 1 - \left ( \frac{6}{16} - \frac{2}{16} \right ) = \\\\\\ 1 - \frac{4}{16} = \\\\\\ 1 - \frac{1}{4} = \\\\\\ \boxed{\frac{3}{4}}$

Nota:

$a^3 + b^3 = (a + b)(a^2 - ab + b^2)$

Espero ter ajudado!

Daniel.