Chapter 5, Problem 1PS

To determine

To identify: The trigonometric functions of $\theta$ in terms of $\cos \theta$ .

Answer to Problem 1PS

The trigonometric functions of $\theta$ in terms of $\cos \theta$ are

  $\begin{array}{c}\sin \theta =\sqrt{1-{\cos }^2\theta }\\ \tan \theta =\frac{\sqrt{1-{\cos }^2\theta }}{\cos \theta }\\ \cot \theta =\frac{\cos \theta }{\sqrt{1-{\cos }^2\theta }}\\ \csc \theta =\frac{1}{\sqrt{1-{\cos }^2\theta }}\\ \sec \theta =\frac{1}{\cos \theta }\end{array}$

Explanation of Solution

Concept Involved:

Trigonometric functions are:

  $\sin \theta ,\cos \theta ,\tan \theta ,\cot \theta ,\sec \theta ,\csc \theta$

Since,

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

Then,

  $\begin{array}{l}{\sin }^2\theta =1-{\cos }^2\theta \\ \Rightarrow \sin \theta =\sqrt{1-{\cos }^2\theta }\end{array}$

Also,

  $\begin{array}{c}\tan \theta =\frac{\sin \theta }{\cos \theta }\\ =\frac{\sqrt{1-{\cos }^2\theta }}{\cos \theta }\end{array}$

And

  $\begin{array}{c}\cot \theta =\frac{1}{\tan \theta }\\ =\frac{\cos \theta }{\sqrt{1-{\cos }^2\theta }}\end{array}$

Now $\csc \theta =\frac{1}{\sin \theta }\text{and}\sec \theta =\frac{1}{\cos \theta }$

Then,

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

Conclusion:

The trigonometric functions of $\theta$ in terms of $\cos \theta$ are

  $\begin{array}{c}\sin \theta =\sqrt{1-{\cos }^2\theta }\\ \tan \theta =\frac{\sqrt{1-{\cos }^2\theta }}{\cos \theta }\\ \cot \theta =\frac{\cos \theta }{\sqrt{1-{\cos }^2\theta }}\\ \csc \theta =\frac{1}{\sqrt{1-{\cos }^2\theta }}\\ \sec \theta =\frac{1}{\cos \theta }\end{array}$