Chapter 4.2, Problem 84E

To determine

Find the function $h(t)=f(t)g(t)$ is even or odd when $\text{f(t)=}\text{sin}\text{t}$ and $\text{g(t)}\text{=}\text{tan}\text{t}$ are odd functions.

Answer to Problem 84E

Even Function

Explanation of Solution

Given: $\text{f(t)=}\text{sin}\text{t}$ and $\text{g(t)}\text{=}\text{tan}\text{t}$ are odd functions.

Concept Used:

Even Functions

A function is "even" when: f(x) = f(-x) for all x.

In other words there is symmetry about the y-axis (like a reflection):

In Trigonometry Cosine and Secant functions are Even Function.

Odd Functions

A function is "odd" when: f(-x) = -f(x) for all x

Note the minus in front of f(x): -f(x), And we get origin symmetry:

In Trigonometry, the sine, cosecant, tangent and cotangent functions are odd functions .

Calculation:

  $\text{h(t)}\text{=}\text{f(t}\text{)}\text{g(}\text{t}\text{)}\text{=}\text{sin}\text{t}\text{tan}\text{t}$

Then $\text{h}(-\text{t})=\text{f}(-\text{t})\text{g}(-\text{t})=\sin (-\text{t})\tan (-\text{t})$

Now $\text{h}(-\text{t})=\sin (-\text{t})\tan (-\text{t})=[-\sin (\text{t})][-\tan (\text{t})]$ ; since $\sin (\text{t});\tan (\text{t})$ are odd functions so $\tan (-\text{t})=-\tan (\text{t});\sin (-\text{t})=-\sin \text{t}$

Then $\text{h}(-\text{t})=\sin (-\text{t})\cos (\text{t})=-\sin (\text{t})\cos (\text{t})$; since $\text{sin}\text{t}$ is an odd function so $\sin (-\text{t})=-\sin \text{t}$

Since $\text{h}(-\text{t})=\sin (\text{t})\tan (\text{t})$; $\text{h}(-\text{t})$is an Even function.

Thus, $h(t)=f(t)g(t)$ is an even function.