Chapter 5, Problem 1RCC

a.

To determine

Explain what is a unit circle.

Answer to Problem 1RCC

The unit circle is a circle of radius 1.

  

Explanation of Solution

The unit circle is the circle of radius 1 centered at the origin (0, 0) in the Cartesian coordinate system at the Euclidean plane.

If (x, y) is a point on the unit circle’s circumference, then $\left|x\right|$ and $\left|y\right|$ are the lengths of the legs of a right triangle whose hypotenuse length is 1. Thus, by Pythagorean theorem x and y satisfy the equation,

  $x^2+y^2=1$

Program:

clc
clear
close all
symsxy
r=1;
f=x^2+y^2-r^2;
s=ezplot(f);
s.LineWidth=1.25;
set(gca,'Linewidth',1.25,'fontsize',12)
axis square
axis([-1 1 -1 1])

Query:

• First, we have defined the given data sets.
• Then using a function “scatter (t, y)” sketch a scatter plot.

b.

To determine

Explain what is terminal point determined by a real number t.

Answer to Problem 1RCC

Let’s take a unit circle, start a walk from the point (1, 0) for a distance of ‘t’ units. The point you end up called terminal point.

  

Explanation of Solution

Calculation:

Let’s take a unit circle,

Then the terminal point would be $(x,y)=(\cos \theta ,\sin \theta )$ ,

Where,

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

Consider $\theta =\frac{\pi }{4}$ ,

Then,

  $\begin{array}{l}x=\cos \left(\frac{\pi }{4}\right)=\frac{\sqrt{2}}{2}\\ y=\sin \left(\frac{\pi }{4}\right)=\frac{\sqrt{2}}{2}\end{array}$

The terminal point is,

  $terminalpoint =\left(\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2}\right)$

Program:

clc
clear
close all
theta=pi/4;
x=cos(theta);
y=sin(theta);
teminalpoint=[x,y];

Query:

• First, we have defined the angle.
• Then calculate the value of x and y.
• Then simplify and get the terminal point.

c.

To determine

Explain what is reference number t’ associated with t.

Answer to Problem 1RCC

The reference number $t\text{'}$ associated with $t$ is the shortest distance along the circle between the terminal point $t$ and the x-axis.

Explanation of Solution

Let’s take,

  $t=\frac{5\pi }{4}$

Then the reference number would be written as,

  $t\text{'}=\frac{5\pi }{4}-\pi$

  $t\text{'}=\frac{\pi }{4}$

  

Conclusion:

The reference number is

  $t\text{'}=\frac{\pi }{4}$

d.

To determine

Show the sign of the trigonometric functions in the different quadrants.

Answer to Problem 1RCC

The solution is,

  

Explanation of Solution

The distance from a terminal point to the origin is always positive, but the signs of coordinates of $x$ and $y$ may be differed (positive or negative). Thus, in the first quadrant, where  $x$ and $y$ coordinates are all positive. So, all six trigonometric functions have positive values. In the second quadrant, only sine and cosecant are positive. In the third quadrant, only tangent and cotangent are positive. Finally, in the fourth quadrant, only cosine and secant are positive. The following diagram may help to understand.