Chapter 12.2, Problem 34WE

To determine

Express the other five trigonometric functions of $\varphi$ in terms of v .

Answer to Problem 34WE

  $\sin \varphi =\sqrt{1-v^2}$

  $\tan \varphi =\frac{\sqrt{1-v^2}}{v}$

  $\csc \varphi =\frac{1}{\sqrt{1-v^2}}$

  $\sec \varphi =\frac{1}{v}$

  $\cot \varphi =\frac{v}{\sqrt{1-v^2}}$

Explanation of Solution

Given:

Angle $\varphi$ is acute and $\cos \varphi =v$ .

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 $\varphi$ is acute , all the six trigonometric angles are positive.

  $\cos \varphi =v$

Since

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

Since

  $\tan \varphi =\frac{\sin \varphi }{\cos \varphi }=\frac{\sqrt{1-v^2}}{v}$

  $\csc \varphi =\frac{1}{\sin \varphi }=\frac{1}{\sqrt{1-v^2}}$

  $\sec \varphi =\frac{1}{\cos \varphi }=\frac{1}{v}$

  $\cot \varphi =\frac{1}{\tan \varphi }=\frac{v}{\sqrt{1-v^2}}$