Respostas

2013-09-04T19:34:08-03:00
Para resolver isso, você precisa de saber:
\sin(a+b)=\sin(a)\cos(b)+\sin(b)\cos(a)

Temos:
\sin(\frac{\pi}{6}-x)
Vamos usar:
a=\frac{\pi}{6}
b=-x

\sin(\frac{\pi}{6}-x)=\sin(\frac{\pi}{6})\cos(-x)+\sin(-x)\cos(\frac{\pi}{6})

Sabemos que \frac{\pi}{6}rad=30^{o} e \sin(30^{o})=\frac{1}{2} e \cos(30^{o})=\frac{\sqrt{3}}{2}. Portanto:
\sin(\frac{\pi}{6}-x)=\frac{1}{2}\cos(-x)+\sin(-x)\frac{\sqrt{3}}{2}

Note que \sin(-x)=-\sin(x) e \cos(-x)=\cos(x). Portanto,
\sin(\frac{\pi}{6}-x)=\frac{1}{2}\cos(x)-\sin(x)\frac{\sqrt{3}}{2}

\sin(\frac{\pi}{6}-x)=\frac{1}{2}\cos(x)-\frac{1}{3}\frac{\sqrt{3}}{2}

Agora, só falta encontrarmos o valor de \cos(x).

Como
\sin(a)^2+\cos(a)^2=1

Então:
\sin(x)^2+\cos(x)^2=1

(\frac{1}{3})^2+\cos(x)^2=1

\cos(x)^2=1-\frac{1}{9}

\cos(x)=\sqrt{\frac{8}{9}}=\frac{\sqrt{8}}{3}

Voltando agora:
\sin(\frac{\pi}{6}-x)=\frac{1}{2}\frac{\sqrt{8}}{3}-\frac{1}{3}\frac{\sqrt{3}}{2}

\sin(\frac{\pi}{6}-x)=\frac{\sqrt{8}}{6}-\frac{\sqrt{3}}{6}

\sin(\frac{\pi}{6}-x)=\frac{\sqrt{8}-\sqrt{3}}{6}
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