Question

find the value of the remaining five trigonometric functions of $\vartheta$

role="math" localid="1646316653225" $$\begin{gathered} \cos \vartheta =\frac{2}{3} \\ \frac{3\pi }{2}<\vartheta <2\pi \end{gathered}$$

Short answer

the value of the remaining five trigonometric functions of $\vartheta$are:

$$\begin{gathered} \sin \vartheta =-\frac{\sqrt{5}}{3} \\ \tan \vartheta =-\frac{\sqrt{5}}{2} \\ \cos ec\vartheta =\frac{-3}{\sqrt{5}} \\ sec\vartheta =\frac{3}{2} \\ cot\vartheta =-\frac{2}{\sqrt{5}} \end{gathered}$$

Step-by-step solution

Step 1. Given information

$$\begin{gathered} \cos \vartheta =\frac{2}{3} \\ \frac{3\pi }{2}<\vartheta <2\pi \end{gathered}$$

Step 2. Solving for sinϑ

We know ;

${\sin }^2\vartheta +{\cos }^2\vartheta =1$

Using this :

$$\begin{gathered} \sin \vartheta =\sqrt{1-{\cos }^2\vartheta } \\ \sin \vartheta =\sqrt{1-{\left(\frac{2}{3}\right)}^2} \\ \sin \vartheta =\sqrt{1-\frac{4}{9}} \\ \sin \vartheta =\sqrt{\frac{5}{9}} \\ \sin \vartheta =\frac{\sqrt{5}}{3} \end{gathered}$$

Since theta is in fourth quadrant so,

$\sin \vartheta =-\frac{\sqrt{5}}{3}$

Step 3. Solving for cosecϑ , secϑ , cotϑ

We know ;

$$\begin{gathered} \cos ec\vartheta =\frac{1}{\sin \vartheta } \\ sec\vartheta =\frac{1}{\cos \vartheta } \\ cot\vartheta =\frac{1}{\tan \vartheta } \end{gathered}$$

Using the above identities :

$$\begin{gathered} \cos ec\vartheta =-\frac{3}{\sqrt{5}} \\ sec\vartheta =\frac{3}{2} \\ cot\vartheta =-\frac{2}{\sqrt{5}} \end{gathered}$$