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Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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Use online resources, notes, videos, and group discussions to answer the following questions;

a. What is the definition of the unit circle?

b. Research a way to memorize the unit circle and write about the strategy.

c. where do the points around the unit circle come from? Explain.

Expert Solution

Step 1

The unit circle is a circle of radius 1 and center (0,0) in the Cartesian Coordinate System.

The angles in radians are marked on the unit circle as shown below.

Note that, each point along the circumference of the circle corresponds to a cosine and sine of the angle from the positive x-axis, called $θ$ , and can be listed as $(\cos θ, \sin θ)$ .

The easiest way to remember the angles and points on the unit circle is given as follows.

It is known that $\frac{π}{6} = 30 °$ . Consider going around the circle with an increment of 30 degrees. That is, write the integer multiples of $\frac{π}{6} = 30 °$ .

$\frac{0 π}{6} = 0 \frac{1 π}{6} = \frac{π}{6} \frac{2 π}{6} = \frac{π}{3} \frac{3 π}{6} = \frac{π}{2} \frac{4 π}{6} = \frac{2 π}{3} \frac{5 π}{6} = \frac{5 π}{6} \frac{6 π}{6} = π \frac{7 π}{6} = \frac{7 π}{6} \frac{8 π}{6} = \frac{4 π}{3} \frac{9 π}{6} = \frac{3 π}{2} \frac{10 π}{6} = \frac{5 π}{3} \frac{11 π}{6} = \frac{11 π}{6} \frac{12 π}{6} = 2 π$

Now, imagine a clock as shown below.

Now, let $\frac{π}{6} = 30 °$ corresponds to 5 minutes. Start from 3 by labelling 0 and go counter clock wise by marking $\frac{π}{6}, \frac{π}{3}, \frac{π}{2}, \frac{2 π}{3}, \frac{5 π}{6}, π, \frac{7 π}{6}, \frac{4 π}{3}, etc$ for 2,1,12,11,10,9,8,7 etc. Then, 3 will correspond to 0 and $2 π$ .

Note that, each point along the circumference of the circle can be listed as $(\cos θ, \sin θ)$ .

Since $\cos 0 = 1 and \sin 0 = 0$ , 3 in the clock will correspond to (0,1).

Similarly, since $\cos \frac{π}{6} = \frac{\sqrt{3}}{2} and \cos \frac{π}{6} = \frac{1}{2}$ , 2 in the clock will correspond to $(\frac{\sqrt{3}}{2}, \frac{1}{2})$ and so on. Note that, the 10 in the clock corresponds to $(- \frac{\sqrt{3}}{2}, \frac{1}{2})$ , 8 corresponds to $(- \frac{\sqrt{3}}{2}, - \frac{1}{2})$ and 4 corresponds to $(\frac{\sqrt{3}}{2}, - \frac{1}{2})$ . That is, on point in the first quadrant is obtained, then the corresponding reflection points in the other quadrants can be easily found by using the following table.

Quadrant	Sign
I	(+,+)
II	(-,+)
III	(-,-)
IV	(+,-)

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