Alguns cálculos auxiliares:
[tex3]\circ \sen^3(x-270^{\circ}) = \cos ^3x \\\\
\circ \cos (360^{\circ} - x) = \cos x \\\\
\circ \tan^3 (x-90^{\circ}) = \frac{(-\cos x)^3}{\sen^3x} \\\\
\circ \cos^3 (x-270^{\circ}) = -\sen^3 x[/tex3]
Substituindo:
[tex3]f(x) = \frac{\cos^3x \cdot \cos x }{\frac{-\cos^3 x}{\cancel{\sen^3x}} \cdot \cancelto{-1}{-\sen^3 x}} \hspace{10mm} \therefore \\\\ f(x) = \frac{\cos ^3x \cdot \cos x}{\cos^3 x}, \cos ^3x \neq 0 \Leftrightarrow x \neq \frac{k\pi}{2}, k \in \mathbb{Z} : f(x) = \cos x[/tex3]
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Problema 11
(ITA-2004) Considerando as funções
[tex3]\text{arc} \,\, \sen :[-1;+1] \rightarrow \left [ -\frac{\pi}{2}, \frac{\pi}{2} \right ][/tex3]
e
[tex3]\text{arc} \,\, \sen :[-1;+1] \rightarrow \left[ -\frac{\pi}{2} , \frac{\pi}{2} \right][/tex3]
assinale o valor de [tex3]\cos \left( \text{arc} \sen \frac{3}{5} + \text{arc} \cos \frac{4}{5} \right)[/tex3]
a) [tex3]\frac{6}{25}[/tex3]
b) [tex3]\frac{7}{25}[/tex3]
c) [tex3]\frac{1}{3}[/tex3]
d) [tex3]\frac{2}{5}[/tex3]
e) [tex3]\frac{5}{12}[/tex3]
Alternativa b