Chapter 1.6, Problem 45E

(a)

To determine

To show: Thefunction $\cot x$ is an odd function of $x$ .

Explanation of Solution

Given information:The given function is $\cot x$ .

Proof:

Simplify the given function

  $\cot x=\frac{\cos x}{\sin x}$

The cosine is an even function and the sine is an odd function.

  $\begin{array}{c}\cot \left(-x\right)=\frac{\cos \left(-x\right)}{\sin \left(-x\right)}\\ =-\frac{\cos x}{\sin x}\\ =-\cot x\end{array}$

As $\cot \left(-x\right)=-\cot x$ , the cotangent is an odd function.

Hence, it is proved that $\cot x$ is an odd function of $x$ .

(b)

To determine

To show: The quotient of an even function and an odd function is an odd function.

Explanation of Solution

Proof:

A function $f$ is said to be an even function if $f\left(-x\right)=f\left(x\right)$ for all $x$ in its domain.

A function $f$ is said to be odd function if $f\left(-x\right)=-f\left(x\right)$ for all $x$ in its domain

Assume that $f\left(x\right)$ is even and $g\left(x\right)$ is odd function.

To check that quotient of an even function and an odd function is an odd function, substitute $-x$ for $x$ in the function.

  $\begin{array}{c}\frac{f\left(-x\right)}{g\left(-x\right)}=\frac{f\left(x\right)}{-g\left(x\right)}\\ =-\frac{f\left(x\right)}{g\left(x\right)}\end{array}$

Hence, it is proved that the quotient of an even function and an odd function is an odd function.