Question

find the remaining five trigonometric functions of ( \theta ).

( sin \theta=\frac{2}{3}, \theta ) in quadrant ii

complete the following table.

( sin \theta=\frac{2}{3} )
( 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 \(\sin\theta = \frac{2}{3}\), then \(\csc\theta=\frac{1}{\frac{2}{3}}=\frac{3}{2}\).

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 \(\sin\theta=\frac{2}{3}\), we get \(\cos\theta=-\sqrt{1 - (\frac{2}{3})^{2}}=-\sqrt{1-\frac{4}{9}}=-\sqrt{\frac{5}{9}}=-\frac{\sqrt{5}}{3}\).

Step3: Find \(\sec\theta\)

Since \(\sec\theta=\frac{1}{\cos\theta}\), and \(\cos\theta=-\frac{\sqrt{5}}{3}\), then \(\sec\theta=\frac{1}{-\frac{\sqrt{5}}{3}}=-\frac{3}{\sqrt{5}}=-\frac{3\sqrt{5}}{5}\).

Step4: Find \(\tan\theta\)

Since \(\tan\theta=\frac{\sin\theta}{\cos\theta}\), substitute \(\sin\theta=\frac{2}{3}\) and \(\cos\theta=-\frac{\sqrt{5}}{3}\), we get \(\tan\theta=\frac{\frac{2}{3}}{-\frac{\sqrt{5}}{3}}=-\frac{2}{\sqrt{5}}=-\frac{2\sqrt{5}}{5}\).

Step5: Find \(\cot\theta\)

Since \(\cot\theta=\frac{1}{\tan\theta}\), and \(\tan\theta=-\frac{2\sqrt{5}}{5}\), then \(\cot\theta=\frac{1}{-\frac{2\sqrt{5}}{5}}=-\frac{5}{2\sqrt{5}}=-\frac{\sqrt{5}}{2}\).

Answer:

\(\cos\theta =-\frac{\sqrt{5}}{3}\), \(\tan\theta =-\frac{2\sqrt{5}}{5}\), \(\csc\theta=\frac{3}{2}\), \(\sec\theta =-\frac{3\sqrt{5}}{5}\), \(\cot\theta =-\frac{\sqrt{5}}{2}\)