Question

find the remaining five trigonometric functions of \\( \theta \\).
\\( \sin \theta = \frac { 5 } { 6 }, \theta \\) in quadrant ii
complete the following table.
\\( \sin \theta = \frac { 5 } { 6 } \\)
\\( \cos \theta = \\)
\\( \tan \theta = \\)
\\( \csc \theta = \\)
\\( \sec \theta = \\)
\\( \cot \theta = \\)
(simplify your answer, including any radicals. use integers or fractions for any numbers in the expression.)

Explanation:

Step1: Find \(\csc\theta\)

Since \(\csc\theta=\frac{1}{\sin\theta}\), and \(\sin\theta = \frac{5}{6}\), then \(\csc\theta=\frac{6}{5}\)

Step2: Find \(\cos\theta\)

Use the identity \(\sin^{2}\theta+\cos^{2}\theta = 1\). So \(\cos^{2}\theta=1-\sin^{2}\theta\). Substitute \(\sin\theta=\frac{5}{6}\), we get \(\cos^{2}\theta=1 - (\frac{5}{6})^{2}=1-\frac{25}{36}=\frac{36 - 25}{36}=\frac{11}{36}\). Since \(\theta\) is in quadrant II, \(\cos\theta<0\). Then \(\cos\theta=-\frac{\sqrt{11}}{6}\)

Step3: Find \(\sec\theta\)

Since \(\sec\theta=\frac{1}{\cos\theta}\), and \(\cos\theta=-\frac{\sqrt{11}}{6}\), then \(\sec\theta=-\frac{6}{\sqrt{11}}=-\frac{6\sqrt{11}}{11}\)

Step4: Find \(\tan\theta\)

Since \(\tan\theta=\frac{\sin\theta}{\cos\theta}\), substitute \(\sin\theta=\frac{5}{6}\) and \(\cos\theta=-\frac{\sqrt{11}}{6}\), we get \(\tan\theta=\frac{\frac{5}{6}}{-\frac{\sqrt{11}}{6}}=-\frac{5}{\sqrt{11}}=-\frac{5\sqrt{11}}{11}\)

Step5: Find \(\cot\theta\)

Since \(\cot\theta=\frac{1}{\tan\theta}\), and \(\tan\theta=-\frac{5\sqrt{11}}{11}\), then \(\cot\theta=-\frac{\sqrt{11}}{5}\)

Answer:

\(\cos\theta=-\frac{\sqrt{11}}{6}\), \(\tan\theta =-\frac{5\sqrt{11}}{11}\), \(\csc\theta=\frac{6}{5}\), \(\sec\theta=-\frac{6\sqrt{11}}{11}\), \(\cot\theta=-\frac{\sqrt{11}}{5}\)