Trigonometry (10th Edition)

Published by Pearson
ISBN 10: 0321671775
ISBN 13: 978-0-32167-177-6

Chapter 6 - Test: 5b

Answer

$\frac{4\sqrt2}{9}$

Work Step by Step

Let $\theta = \cos^{-1}{(\frac{1}{3})}$ This means that $\cos{\theta}=\frac{1}{3}$ Since $\cos{\theta} = \frac{x}{r}$ and $\cos{\theta}=\frac{1}{3}$, then we can assume that $x=1$ and $r=3$. RECALL: In a unit circle, $r^2=x^2+y^2$. Use the formula above to solve for $y$ to obtain: $r^2=x^2+y^2 \\3^2 = 1^2 + y^2 \\9=1+y^2 \\9-1=y^2 \\8=y^2 \\\sqrt8=\sqrt{y^2} \\\sqrt{4(2)}=y \\2\sqrt2=y$ Since $\sin{\theta} = \frac{y}{r}$, then $\sin{\theta} = \frac{2\sqrt2}{3}$ With $\theta=\cos^{-1}{(\frac{1}{3})}$, then: $\sin{\left(2\cos^{-1}{\frac{1}{3}}\right)}=\sin{2\theta}$ RECALL: $\sin{2\theta} = 2\sin{\theta}\cos{\theta}$ Thus, $\sin{\left(2\cos^{-1}{\frac{1}{3}}\right)} \\=\sin{2\theta} \\= 2\sin{\theta}\cos{\theta} \\=2(\frac{2\sqrt2}{3})(\frac{1}{3}) \\=\frac{4\sqrt2}{9}$
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