Chapter 5.1, Problem 116E

To determine

To express: Each trigonometric function of $\theta$ in terms of $\cos \theta$ .

Answer to Problem 116E

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

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

Explanation of Solution

Given information:

Each trigonometric function of $\theta$ in terms of $\cos \theta$ .

Formula used:

The Pythagorean identity:

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

Calculation:

As per problem,

• To express $\sin \theta$ in terms of $\cos \theta$,

From the Pythagorean identity: ${\sin }^2\theta +{\cos }^2\theta =1$ , rearrange to find $\sin \theta$

  $\begin{array}{l}{\sin }^2\theta =1-{\cos }^2\theta \text{ [Rearrange }{\sin }^2\theta +{\cos }^2\theta =1]\\ \sin \theta =\pm \sqrt{1-{\cos }^2\theta }\text{ [Take square root of both sides]}\end{array}$

• To express $\text{tan}\theta$ in terms of $\cos \theta$,

  $\begin{array}{l}\text{tan}\theta =\frac{\sin \theta }{\cos \theta }\\ =\frac{\pm \sqrt{1-{\cos }^2\theta }}{\cos \theta }\text{ [Write }\sin \theta \mathrm{in}\text{ terms of }\cos \theta \text{ as obtained above]}\end{array}$

• To express $\text{csc}\theta$ in terms of $\cos \theta$,

  $\begin{array}{l}\text{csc}\theta \text{=}\frac{1}{\sin \theta }\text{ [Reciprocal identity]}\\ \text{ =}\frac{1}{\pm \sqrt{1-{\cos }^2\theta }}\text{ [Write }\sin \theta \mathrm{in}\text{ terms of cos}\theta \text{ as obtained above]}\end{array}$

• To express $\text{sec}\theta$ in terms of $\cos \theta$,

  $\text{sec}\theta \text{=}\frac{1}{\cos \theta }\text{ [Reciprocal identity]}$

• To express $\text{cot}\theta$ in terms of $\cos \theta$,

  $\begin{array}{l}\text{cot}\theta \text{=}\frac{\cos \theta }{\sin \theta }\text{ [Quotient identity]}\\ \text{ =}\frac{\cos \theta }{\pm \sqrt{1-{\cos }^2\theta }}\text{ [Write }\sin \theta \mathrm{in}\text{ terms of }\cos \theta \text{ as obtained above]}\end{array}$

Hence, the trigonometric functions of $\theta$ in terms of $\cos \theta$ are:

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