Chapter 5, Problem 14RE

To determine

To evaluate: the remaining trigonometric function of the angle..

Answer to Problem 14RE

  $\begin{array}{l}\sin \theta =\frac{2\sqrt{2}}{2},\cos \theta =\frac{1}{3},\tan \theta =2\sqrt{2}\\ \csc \theta =\frac{3\sqrt{2}}{4},\sec \theta =3,\cot \theta =\frac{\sqrt{2}}{4}\end{array}$

Explanation of Solution

Given information:

Given trigonometric function is

  $\csc \left(\frac{\pi }{2}-\theta \right)=-3,\sin \theta =\frac{2\sqrt{2}}{3}$ .

Calculation:

Use the co function identity, $\sec \theta =\csc \left(\frac{\pi }{2}-\theta \right)$ , substitute 3 for $\csc \left(\frac{\pi }{2}-\theta \right)$ in the identity.

  $\sec \theta =3$

Therefore, the value of $\sec \theta$ is 3.

Now apply the reciprocal identity, $\cos \theta =\frac{1}{\sec \theta }$ . Substitute 3 for $\sec \theta$ in the identity.

  $\cos \theta =\frac{1}{3}$

Therefore, the value of $\cos \theta$ is $\frac{1}{3}$ .

Now, apply the quotient identity, $\tan \theta =\frac{\sin \theta }{\cos \theta }$ . Substitute $\frac{2\sqrt{2}}{3}$ for $\sin \theta$ , $\frac{1}{3}$ for $\cos \theta$ and solve for $\tan \theta$ .

  $\begin{array}{c}\tan \theta =\frac{\frac{2\sqrt{2}}{3}}{\frac{1}{3}}\\ =2\sqrt{2}\end{array}$

Therefore, the value of $\tan \theta$ is $2\sqrt{2}$ .

Find the remaining trigonometric identities using the reciprocal identities,

  $\csc \theta =\frac{1}{\sin \theta },\sec \theta =\frac{1}{\cos \theta },\cot \theta =\frac{1}{\tan \theta }$

Replace $\sin \theta$ with $\frac{2\sqrt{2}}{3}$ and $\tan \theta$ with $2\sqrt{2}$ , in the identities.

  $\begin{array}{l}\csc \theta =\frac{1}{\sin \theta }=\frac{1}{\frac{2\sqrt{2}}{3}}=\frac{3\sqrt{2}}{4}\\ \cot \theta =\frac{1}{\tan \theta }=\frac{1}{2\sqrt{2}}=\frac{\sqrt{2}}{4}\end{array}$

Therefore, the six trigonometric identities are

  $\begin{array}{l}\sin \theta =\frac{2\sqrt{2}}{2},\cos \theta =\frac{1}{3},\tan \theta =2\sqrt{2}\\ \csc \theta =\frac{3\sqrt{2}}{4},\sec \theta =3,\cot \theta =\frac{\sqrt{2}}{4}\end{array}$