Chapter 4.4, Problem 7E

To determine

To fill: in the blank of the given statement.

Answer to Problem 7E

  $r=\sqrt{x^2+y^2}\ne 0$ and the sine and the cosine functions are defined for all real values of $\theta$ , if $r>0$ .

Explanation of Solution

Given information:

Because $r=\sqrt{x^2+y^2}$ cannot be -------, the $\sin e$ and $\cos ine$ functions are -----------for any real value of $\theta$ .

Calculation:

As $r=\sqrt{x^2+y^2}$ cannot be zero, the sine and cosine functions are defined for any real value of $\theta$ .

Let us consider that $\theta$ be an angle in standard position with $\left(x,y\right)$ as shown in figure.

Also we have

  $\begin{array}{c}r=\sqrt{x^2+y^2}\\ \ne 0\end{array}$

If $r$ will be zero, only a point will exist and no angle will be actually made, hence $r>0$ .

Since, we have

  $\begin{array}{l}\sin \theta =\frac{y}{r}\\ \cos \theta =\frac{x}{r}\end{array}$

Hence, the sine and the cosine functions are defined for all real values of $\theta$ , if $r>0$ .