Question

chapter 7 quiz
150 points possible answered: 11/15
question 12
if \\( \sin x = \frac { 1 } { 6 } \\), \\( x \\) in quadrant i, then find (without finding \\( x \\) )
\\( \sin ( 2 x ) = \\)
\\( \cos ( 2 x ) = \\)
\\( \tan ( 2 x ) = \\)

Explanation:

Step1: Find \(\cos x\)

Use the identity \(\sin^{2}x+\cos^{2}x = 1\).
Given \(\sin x=\frac{1}{6}\), then \(\cos^{2}x=1-\sin^{2}x=1 - (\frac{1}{6})^{2}=1-\frac{1}{36}=\frac{35}{36}\).
Since \(x\) is in quadrant I, \(\cos x=\frac{\sqrt{35}}{6}\).

Step2: Find \(\sin(2x)\)

Use the double - angle formula \(\sin(2x)=2\sin x\cos x\).
Substitute \(\sin x=\frac{1}{6}\) and \(\cos x=\frac{\sqrt{35}}{6}\) into the formula:
\(\sin(2x)=2\times\frac{1}{6}\times\frac{\sqrt{35}}{6}=\frac{\sqrt{35}}{18}\).

Step3: Find \(\cos(2x)\)

Use the double - angle formula \(\cos(2x)=1 - 2\sin^{2}x\).
Substitute \(\sin x=\frac{1}{6}\) into the formula:
\(\cos(2x)=1-2\times(\frac{1}{6})^{2}=1-\frac{2}{36}=\frac{17}{18}\).

Step4: Find \(\tan(2x)\)

Use the formula \(\tan(2x)=\frac{\sin(2x)}{\cos(2x)}\).
Substitute \(\sin(2x)=\frac{\sqrt{35}}{18}\) and \(\cos(2x)=\frac{17}{18}\) into the formula:
\(\tan(2x)=\frac{\frac{\sqrt{35}}{18}}{\frac{17}{18}}=\frac{\sqrt{35}}{17}\).

Answer:

\(\sin(2x)=\frac{\sqrt{35}}{18}\), \(\cos(2x)=\frac{17}{18}\), \(\tan(2x)=\frac{\sqrt{35}}{17}\)