Question

Blake is told that an angle has a cosine of -(1)/(√(6)) and is in Quadrant III. Use the Pythagorean identity \sin^2(\theta) + \cos^2(\theta) = 1, the trigonometric identity \tan \theta = (\sin \theta)/(\cos \theta), and the quadrant information to find the tangent.

Answer 1

To find the tangent of the angle θ given that \(\cos \theta = -\frac{1}{\sqrt{6}}\) and that θ is in Quadrant III, we will use the Pythagorean identity and relevant trigonometric definitions.

1. **Use the Pythagorean Identity to find \(\sin \theta\)**:
According to the Pythagorean identity, we have:
\[
\sin^2(\theta) + \cos^2(\theta) = 1
\]
Plugging in the value of \(\cos \theta\):
\[
\sin^2(\theta) + \left(-\frac{1}{\sqrt{6}}\right)^2 = 1
\]
Simplify the square of cosine:
\[
\sin^2(\theta) + \frac{1}{6} = 1
\]
Rearranging gives:
\[
\sin^2(\theta) = 1 - \frac{1}{6}
\]
\[
\sin^2(\theta) = \frac{6}{6} - \frac{1}{6} = \frac{5}{6}
\]
Taking the square root of both sides, we find:
\[
\sin(\theta) = \pm \sqrt{\frac{5}{6}}
\]
Since θ is in Quadrant III, where sin is negative, we take the negative root:
\[
\sin(\theta) = -\sqrt{\frac{5}{6}} = -\frac{\sqrt{5}}{\sqrt{6}} = -\frac{\sqrt{30}}{6}
\]

2. **Use the definition of tangent to find \(\tan \theta\)**:
The tangent function is defined as:
\[
\tan \theta = \frac{\sin \theta}{\cos \theta}
\]
Substitute the values we found for sin and cos:
\[
\tan \theta = \frac{-\frac{\sqrt{30}}{6}}{-\frac{1}{\sqrt{6}}}
\]
This simplifies to:
\[
\tan \theta = \frac{\sqrt{30}}{6} \cdot \frac{\sqrt{6}}{1} = \frac{\sqrt{30} \cdot \sqrt{6}}{6} = \frac{\sqrt{180}}{6}
\]
Since \(\sqrt{180} = \sqrt{36 \times 5} = 6\sqrt{5}\), we have:
\[
\tan \theta = \frac{6\sqrt{5}}{6} = \sqrt{5}
\]

Thus, the value of \(\tan \theta\) is:
\[
\boxed{\sqrt{5}}
\]