Question

find the remaining five trigonometic functions of \\( \theta \\).
\\( \sin \theta = \frac { 2 } { 5 }, \theta \\) in quadrant ii
complete the following table.
\\( \sin \theta = \frac { 2 } { 5 } \\)
\\( \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 \(\sin\theta = \frac{2}{5}\), then \(\csc\theta=\frac{5}{2}\).

Step2: Find \(\cos\theta\)

Using the identity \(\sin^{2}\theta+\cos^{2}\theta = 1\), so \(\cos\theta=-\sqrt{1-\sin^{2}\theta}\) (because \(\theta\) is in quadrant II, \(\cos\theta<0\)).
Substitute \(\sin\theta=\frac{2}{5}\), we get \(\cos\theta=-\sqrt{1 - (\frac{2}{5})^{2}}=-\sqrt{1-\frac{4}{25}}=-\sqrt{\frac{21}{25}}=-\frac{\sqrt{21}}{5}\).

Step3: Find \(\sec\theta\)

Since \(\sec\theta=\frac{1}{\cos\theta}\), then \(\sec\theta=-\frac{5}{\sqrt{21}}=-\frac{5\sqrt{21}}{21}\).

Step4: Find \(\tan\theta\)

Since \(\tan\theta=\frac{\sin\theta}{\cos\theta}\), substitute \(\sin\theta=\frac{2}{5}\) and \(\cos\theta =-\frac{\sqrt{21}}{5}\), we get \(\tan\theta=\frac{\frac{2}{5}}{-\frac{\sqrt{21}}{5}}=-\frac{2}{\sqrt{21}}=-\frac{2\sqrt{21}}{21}\).

Step5: Find \(\cot\theta\)

Since \(\cot\theta=\frac{1}{\tan\theta}\), then \(\cot\theta=-\frac{\sqrt{21}}{2}\).

Answer:

\(\cos\theta=-\frac{\sqrt{21}}{5}\), \(\tan\theta =-\frac{2\sqrt{21}}{21}\), \(\csc\theta=\frac{5}{2}\), \(\sec\theta=-\frac{5\sqrt{21}}{21}\), \(\cot\theta=-\frac{\sqrt{21}}{2}\)