Write the polar form of a complex number in the form of a+ib

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Homework Help Overview

The discussion revolves around converting a complex number expressed in polar form, specifically 4{cos(13∏/6)+isin(13∏/6)}, into its rectangular form a+ib. Participants are exploring the trigonometric values associated with the angles involved, particularly how to derive the cosine and sine values from radians.

Discussion Character

  • Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants are attempting to understand how to convert the cosine and sine of radians into fractional forms. There are questions about the basic trigonometric values and how they relate to specific angles, particularly 30° and 390°.

Discussion Status

Some participants are providing insights into the relationship between radians and degrees, while others are questioning the foundational knowledge of trigonometric values. There is an ongoing exploration of how to recognize and derive these values without the aid of a calculator.

Contextual Notes

Participants note constraints such as the lack of a scientific calculator during exams and the absence of relevant trigonometric demonstrations in the textbook. This highlights a gap in foundational knowledge that is being addressed through discussion.

thatguythere
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Homework Statement


4{cos(13∏/6)+isin(13∏/6)}
= 4((√3/2)+(i/2))
= 2√3+2i

Homework Equations


The Attempt at a Solution



This is an example from my textbook. The part which I do not understand is how to convert the cos and sin of radians into those fractions. Any help is greatly appreciated.
 
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thatguythere said:

Homework Statement


4{cos(13∏/6)+isin(13∏/6)}
= 4((√3/2)+(i/2))
= 2√3+2i

Homework Equations



The Attempt at a Solution



This is an example from my textbook. The part which I do not understand is how to convert the cos and sin of radians into those fractions. Any help is greatly appreciated.
That's basic trigonometry.
 
Remember to set you calculator to 'radians' instead of 'degrees' if you don't know trigonometry.
 
I understand how to convert between radians and degrees. What I am unsure of is how to get from the radians to those fractions. When I write the exam, I am not allowed a scientific calculator either, so I definitely need to know this. Also, Sammy, while this may be basic to you, it is not something I have ever learned. So, perhaps you could simply point me in the right direction? My textbook does not demonstrate the relevant trigonometry.
 
thatguythere said:
...

Also, Sammy, while this may be basic to you, it is not something I have ever learned. So, perhaps you could simply point me in the right direction? My textbook does not demonstrate the relevant trigonometry.
A complete circle has 360° which is equivalent to 2π radians.

\displaystyle \frac{13\pi}{6}=2\pi+\frac{\pi}{6}

\displaystyle \frac{\pi}{6}\ radians is equivalent to 30 ° .

Do you know the sine and cosine of 390° ?
 
Finding sine and cosine of 390° or 30° is not difficult. Converting sin390°=0.5 into 1/2 is simple as well. However, when it comes to converting cos390°=0.866 into √3/2, I am at a loss.
 
thatguythere said:
Finding sine and cosine of 390° or 30° is not difficult. Converting sin390°=0.5 into 1/2 is simple as well. However, when it comes to converting cos390°=0.866 into √3/2, I am at a loss.

It's not about "converting" 0.866 into ##\frac{\sqrt{3}}{2}##. Rather it's about recognising that ##\cos{\frac{\pi}{6}} = \frac{\sqrt{3}}{2}##. 0.866 is just a decimal approximation of that irrational number.

To do that, you need to "know" the special right triangle with perpendicular sides measuring ##1## and ##\sqrt{3}## and hypotenuse ##2##. You should be able to verify that Pythagoras Theorem holds for this right triangle. The angles in the triangle are (in degrees): 90, 60 and 30 and you can figure out the equivalent radian measure. You should be able to figure out which angle is 60 deg and which is 30 deg, and the sines and cosines of those special angles should become easy to compute exactly.

BTW, that special right triangle is simply half an equilateral triangle of side length 2, bisected along one of its vertical heights. The reason why one of the angles is 30 deg should now become obvious.
 
Draw an equilateral triangle with each side of length 2. Draw a line from one vertex to the midpoint of the other line. That line will also bisect the opposite side and the angle. Since each angle in an equilateral triangle is 60 degrees= [/itex]\pi/3[/itex] radians, you now have two right triangles with angles \pi/3 and \pi/6 radians and with one leg of length 1 and hypotenuse of length 2. From the Pythagorean theorem, it follows that the other leg has length \sqrt{3}.

From that we can say:
sin(\pi/3)= sin(60)= \frac{\sqrt{3}}{2}
sin(\pi/6)= sin(30)= \frac{1}{2}
 

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