Pré-Vestibular ⇒ (UFPB - 1983) Expressão Trigonométrica Tópico resolvido

[ALDRINMOD]

O valor numérico da expressão [tex3]\frac{\sen 135^\circ \cos 15^\circ \sen 270^\circ}{\sen 195^\circ \sen 75^\circ \sec 105^\circ}[/tex3] é:

a) [tex3]{-}\frac{\sqrt{2}}{2}.[/tex3]
b) [tex3]\frac{\sqrt{2}}{2}.[/tex3]
c) [tex3]\frac{1}{2}.[/tex3]
d) [tex3]{-}\frac{1}{2}.[/tex3]
e) [tex3]\frac{\sqrt{3}}{2}.[/tex3]

[adrianotavares]

Olá, Aldrin.

[tex3]\sen 135^\circ= \sen45^\circ= \frac{\sqrt{2}}{2}[/tex3]

[tex3]\cos 15^\circ= \cos( 45^\circ- 30^\circ)= \frac{\sqrt{2}}{2}\cdot \frac{\sqrt{3}}{2}+ \frac{\sqrt{2}}{2}\cdot \frac{1}{2}= \frac{\sqrt{6}+\sqrt{2}}{4}[/tex3]

[tex3]\sen 270^\circ= -1[/tex3]

[tex3]\sen 195^\circ = \sen( 150^\circ+ 45^\circ)=\frac{1}{2}\cdot \frac{\sqrt{2}}{2}+ \frac{\sqrt{2}}{2}\cdot (-\frac{\sqrt{3}}{2})= \frac{\sqrt{2}-\sqrt{6}}{4}[/tex3]

[tex3]\sen 75^\circ = \sen( 30^\circ+ 45^\circ)=\frac{1}{2}\cdot \frac{\sqrt{2}}{2}+ \frac{\sqrt{2}}{2}\cdot \frac{\sqrt{3}}{2}= \frac{\sqrt{6}+\sqrt{2}}{4}[/tex3]

[tex3]\cos 105^\circ = \cos( 60^\circ+ 45^\circ)= \frac{\sqrt{1}}{2}\cdot \frac{\sqrt{2}}{2}- \frac{\sqrt{3}}{2}\cdot \frac{\sqrt{2}}{2}= \frac{\sqrt{2}-\sqrt{6}}{4}[/tex3]

[tex3]sec 105^\circ = \frac{1}{\cos 105^\circ}= \frac{4}{\sqrt{2}-\sqrt{6}}[/tex3]

Cálculo do valor da expessão:

[tex3]\frac{\(\frac{\sqrt{2}}{2}\)\cdot \(\frac{\sqrt{6}+\sqrt{2}}{4}\)\cdot (-1)}{\(\frac{\sqrt{2}-\sqrt{6}}{4}\)\cdot \(\frac{\sqrt{6}+\sqrt{2}}{4}\).\(\frac{4}{\sqrt{2}-\sqrt{6}}\)} = \frac{-\sqrt{2}}{2}[/tex3]

Alternativa: a