Chapter 12.2, Problem 33WE

To determine

Express the other five trigonometric functions of $\theta$ in terms of u .

Answer to Problem 33WE

  $\cos \theta =\sqrt{1-u^2}$

  $\tan \theta =\frac{u}{\sqrt{1-u^2}}$

  $\csc \theta =\frac{1}{u}$

  $\sec \theta =\frac{1}{\sqrt{1-u^2}}$

  $\cot \theta =\frac{\sqrt{1-u^2}}{u}$

Explanation of Solution

Given:

Angle $\theta$ is acute and $\sin \theta =u$ .

Formula Used:

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

  $\tan \theta =\frac{\sin \theta }{\cos \theta }$

  $\csc \theta =\frac{1}{\sin \theta }$

  $\sec \theta =\frac{1}{\cos \theta }$

  $\cot \theta =\frac{1}{\tan \theta }$

Calculation:

Since angle $\theta$ is acute , all the six trigonometric angles are positive.

  $\sin \theta =u$

Since

  $\begin{array}{l}{\sin }^2\theta +{\cos }^2\theta =1\\ \text{⇒}\cos \theta =\sqrt{1-{\sin }^2\theta }\\ \text{⇒}\cos \theta =\sqrt{1-u^2}\mathrm{..............}[\sin \theta =u]\end{array}$

Since

  $\tan \theta =\frac{\sin \theta }{\cos \theta }=\frac{u}{\sqrt{1-u^2}}$

  $\csc \theta =\frac{1}{\sin \theta }=\frac{1}{u}$

  $\sec \theta =\frac{1}{\cos \theta }=\frac{1}{\sqrt{1-u^2}}$

  $\cot \theta =\frac{1}{\tan \theta }=\frac{\sqrt{1-u^2}}{u}$