anonymous
  • anonymous
Can someone explain how to express -4 in trigonometric form? Really lost. Thank you!
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
jim_thompson5910
  • jim_thompson5910
Polar form: \(\Large (r,\theta)\) Trig form: \(\Large r[\cos(\theta) + i*\sin(\theta)]\) The two forms are very similar
jim_thompson5910
  • jim_thompson5910
write -4 as -4+0i that is in the form a+bi
anonymous
  • anonymous
Ok.

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jim_thompson5910
  • jim_thompson5910
a = -4 b = 0 plug those values into the formulas r = sqrt(a^2 + b^2) theta = arctan(b/a)
anonymous
  • anonymous
So r = 4 and theta = 0?
jim_thompson5910
  • jim_thompson5910
r = 4 is correct but theta = 0 is not
jim_thompson5910
  • jim_thompson5910
|dw:1437696080574:dw|
anonymous
  • anonymous
Isn't 0/-4 equal to 0, and the arctan(0) just 0?
jim_thompson5910
  • jim_thompson5910
arctan(b/a) = arctan(0/(-4)) = 0 is true but we're in the wrong spot add on 180 degrees |dw:1437696139033:dw|
anonymous
  • anonymous
So is r = 4 and theta = -4...?
jim_thompson5910
  • jim_thompson5910
theta = 180 degrees = pi radians
anonymous
  • anonymous
so theta = pi ?
jim_thompson5910
  • jim_thompson5910
yeah
anonymous
  • anonymous
Ok so let me try and see if I can get the answer.
anonymous
  • anonymous
A. 4(cos0degrees + i sin270degrees) B. 4(cos90degrees + i sin90degrees) C. 4(cos0degrees + i sin0degrees) D. 4(cos180degrees + i sin180degrees) Would it be D?
jim_thompson5910
  • jim_thompson5910
The trig form, in radians, is \(\Large 4[\cos(\pi) + i*\sin(\pi)]\) that's equivalent to \(\Large 4[\cos(180^{\circ}) + i*\sin(180^{\circ})]\)
jim_thompson5910
  • jim_thompson5910
you are correct
anonymous
  • anonymous
Thank you! I'll Medal you (:
jim_thompson5910
  • jim_thompson5910
you're welcome
anonymous
  • anonymous
Could you help me do the same thing for -6i?
anonymous
  • anonymous
@jim_thompson5910
jim_thompson5910
  • jim_thompson5910
-6i = 0 + (-6i) now a = 0 and b = -6
jim_thompson5910
  • jim_thompson5910
use the formulas r = sqrt(a^2 + b^2) theta = arctan(b/a)
jim_thompson5910
  • jim_thompson5910
draw a graph to make sure you get the theta in the right spot
anonymous
  • anonymous
So would theta = pi?
jim_thompson5910
  • jim_thompson5910
what is the value of b/a this time?
anonymous
  • anonymous
Undefined actually
jim_thompson5910
  • jim_thompson5910
where is tan(theta) undefined?
anonymous
  • anonymous
Isn't it at 90 degrees and 270 degrees?
jim_thompson5910
  • jim_thompson5910
correct
jim_thompson5910
  • jim_thompson5910
|dw:1437697203957:dw|
jim_thompson5910
  • jim_thompson5910
|dw:1437697245770:dw|
anonymous
  • anonymous
So theta = 3pi/2
jim_thompson5910
  • jim_thompson5910
yep
anonymous
  • anonymous
Therefore making the answer: 6(cos270degrees + i sin270degrees)
jim_thompson5910
  • jim_thompson5910
correct
anonymous
  • anonymous
Thank you so much, could you help me with just one more? I think I understand how to do a single number, but am not sure how to do an equation?
jim_thompson5910
  • jim_thompson5910
what's the question
anonymous
  • anonymous
3-3i
jim_thompson5910
  • jim_thompson5910
that's not an equation. An equation needs an = sign
anonymous
  • anonymous
Sorry, meant expression.
jim_thompson5910
  • jim_thompson5910
compare a+bi with 3-3i to see that a = 3 and b = -3
anonymous
  • anonymous
Oh ok! So then r = 3SqRt2 and theta = -1 ... but I'm still confused as to how you find the theta degree value.
jim_thompson5910
  • jim_thompson5910
theta is not -1
jim_thompson5910
  • jim_thompson5910
b/a = -1, that is true
jim_thompson5910
  • jim_thompson5910
arctan(b/a) = arctan(-1) = ??
anonymous
  • anonymous
-45...
anonymous
  • anonymous
So 315 degrees?
jim_thompson5910
  • jim_thompson5910
yes
anonymous
  • anonymous
So would the answer be: 3SqRt2(cos7pi/4 + i sin7pi/4) ?
jim_thompson5910
  • jim_thompson5910
correct
anonymous
  • anonymous
Thank you so much. I finally understand this!! (: You really helped me understand this, if I could give you more medals I definitely would.
jim_thompson5910
  • jim_thompson5910
that would be the radian form
jim_thompson5910
  • jim_thompson5910
you're welcome
anonymous
  • anonymous
Yes, that is what the answers are listed as.

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