<p>Been sick, just went back to school, and turns out I have a test tomorrow. Im trying to teach my self the material, but im a little confused</p>
<p>here is a sample problem from my text book, but I dont understand how the textbook solved it:</p>
<p>directions: "write the products & quotients in Polar Form, then rectangular"</p>
<p>first of all I dont understand the conceptualdifference between Polar, Rectangular, and Normal.</p>
<p>ny way,</p>
<p>heres the problem:</p>
<h1>4.</h1>
<p>-4(cos pi + i sin pi) * 5(cos (5pi/3) + i sin (5pi/3) )</p>
<p>isnt this already in polar, how do I find the product in polar, something tells me I cant just use the "FOIL" method</p>
<p>and then rectangular?</p>
<p>isn't polar when you eliminate all trigonometrical functions and instead replace them in terms of x and y? I¡'m not really sure though</p>
<p>Simplify, use DeMoivre's.</p>
<p>-4(cos pi + i sin pi) * 5(cos (5pi/3) + i sin (5pi/3) )</p>
<p>= (-4)(5) cis (pi + 5pi/3)
= -20 cis (8pi/3)
= -20 cis (2pi/3)</p>
<p>I see il bandito, but I don't understand that last step where Im assuming you went from polar to rectangular, sorry for my ignorance though</p>
<p>-20cis(8pi/3) -> -20 cis (2pi/3)</p>
<p>No, I subtracted out 6pi/3, aka 2pi, aka a complete rotation about the origin in the argand plane.</p>
<p>oh of course, duh</p>
<p>1 last question,</p>
<p><a href="http://www.sparknotes.com/math/trigonometry/trigonometricequations/section1.html%5B/url%5D">http://www.sparknotes.com/math/trigonometry/trigonometricequations/section1.html</a></p >
<p>Here I understand how they got cosx = .5, & I also see how they got pi/3 (60*) but how did they get 5pi/3. We've done that in class, and I never got it, and never got around to asking my instructer. I know that there is some sort of rule where you add or subtract pi depending on quadrant or something?</p>
sneezy
June 7, 2006, 10:51pm
8
<p>You use your central angle and see which quantrants they apply it to.</p>
<p>Quad 1 is positive for both x, y
Quad 2: x is negative, y is positive
Quad 3: both negative
Quad 4: x positive, y negative</p>
<p>This is, if I understand what you're asking.</p>
<p>Did you ask for an extension? And I have no idea what you're talking about.</p>