Chapter 4, Problem 39RE

(a)

To determine

Find the exact value of each indicated trigonometric function.

Answer to Problem 39RE

  $\csc \theta =3$

Explanation of Solution

Given:

The trigonometric function

  $\sin \theta =\frac{1}{3}$

since

  $\sin \theta =\frac{\text{ Perpendicular }}{\text{ Hypotenuse }}$

  $\csc \theta$

  $\begin{array}{cc}\text{base}& =\sqrt{3^2-1^2}\\ & =\sqrt{9-1}\\ & =\sqrt{8}\\ & =2\sqrt{2}\end{array}$

  $\begin{array}{cc}\csc \theta & =\frac{\text{Hypotenuse}}{\text{Perpendicular}}\\ & =3\end{array}$

(b)

To determine

Find the exact value of each indicated trigonometric function.

Answer to Problem 39RE

  $\cos \theta =\frac{2\sqrt{2}}{3}$

Explanation of Solution

Given:

The trigonometric function

  $\sin \theta =\frac{1}{3}$

since

  $\sin \theta =\frac{\text{ Perpendicular }}{\text{ Hypotenuse }}$

cos θ

  $\begin{array}{cc}\text{base}& =\sqrt{3^2-1^2}\\ & =\sqrt{9-1}\\ & =\sqrt{8}\\ & =2\sqrt{2}\end{array}$

  $\cos \theta =\frac{2\sqrt{2}}{3}$

(c)

To determine

Find the exact value of each indicated trigonometric function.

Answer to Problem 39RE

  $\sec \theta =\frac{3}{2\sqrt{2}}$

Explanation of Solution

Given:

The trigonometric function

  $\sin \theta =\frac{1}{3}$

since

  $\sin \theta =\frac{\text{ Perpendicular }}{\text{ Hypotenuse }}$

sec θ

  $\begin{array}{cc}\text{base}& =\sqrt{3^2-1^2}\\ & =\sqrt{9-1}\\ & =\sqrt{8}\\ & =2\sqrt{2}\end{array}$

  $\sec \theta =\frac{3}{2\sqrt{2}}$

(d)

To determine

Find the exact value of each indicated trigonometric function.

Answer to Problem 39RE

  $\tan \theta =\frac{1}{2\sqrt{2}}$

Explanation of Solution

Given:

The trigonometric function

  $\sin \theta =\frac{1}{3}$

since

  $\sin \theta =\frac{\text{ Perpendicular }}{\text{ Hypotenuse }}$

tan θ

  $\begin{array}{cc}\text{base}& =\sqrt{3^2-1^2}\\ & =\sqrt{9-1}\\ & =\sqrt{8}\\ & =2\sqrt{2}\end{array}$

  $\tan \theta =\frac{1}{2\sqrt{2}}$