Can someone explain how to express -4 in trigonometric form? Really lost. Thank you!

- anonymous

Can someone explain how to express -4 in trigonometric form? Really lost. Thank you!

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- schrodinger

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- jim_thompson5910

Polar form: \(\Large (r,\theta)\)
Trig form: \(\Large r[\cos(\theta) + i*\sin(\theta)]\)
The two forms are very similar

- jim_thompson5910

write -4 as -4+0i
that is in the form a+bi

- anonymous

Ok.

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## More answers

- jim_thompson5910

a = -4
b = 0
plug those values into the formulas
r = sqrt(a^2 + b^2)
theta = arctan(b/a)

- anonymous

So r = 4 and theta = 0?

- jim_thompson5910

r = 4 is correct
but theta = 0 is not

- jim_thompson5910

|dw:1437696080574:dw|

- anonymous

Isn't 0/-4 equal to 0, and the arctan(0) just 0?

- 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

So is r = 4 and theta = -4...?

- jim_thompson5910

theta = 180 degrees = pi radians

- anonymous

so theta = pi ?

- jim_thompson5910

yeah

- anonymous

Ok so let me try and see if I can get the answer.

- 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

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

you are correct

- anonymous

Thank you! I'll Medal you (:

- jim_thompson5910

you're welcome

- anonymous

Could you help me do the same thing for -6i?

- anonymous

@jim_thompson5910

- jim_thompson5910

-6i = 0 + (-6i)
now a = 0 and b = -6

- jim_thompson5910

use the formulas
r = sqrt(a^2 + b^2)
theta = arctan(b/a)

- jim_thompson5910

draw a graph to make sure you get the theta in the right spot

- anonymous

So would theta = pi?

- jim_thompson5910

what is the value of b/a this time?

- anonymous

Undefined actually

- jim_thompson5910

where is tan(theta) undefined?

- anonymous

Isn't it at 90 degrees and 270 degrees?

- jim_thompson5910

correct

- jim_thompson5910

|dw:1437697203957:dw|

- jim_thompson5910

|dw:1437697245770:dw|

- anonymous

So theta = 3pi/2

- jim_thompson5910

yep

- anonymous

Therefore making the answer:
6(cos270degrees + i sin270degrees)

- jim_thompson5910

correct

- 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

what's the question

- anonymous

3-3i

- jim_thompson5910

that's not an equation. An equation needs an = sign

- anonymous

Sorry, meant expression.

- jim_thompson5910

compare a+bi with 3-3i to see that a = 3 and b = -3

- 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

theta is not -1

- jim_thompson5910

b/a = -1, that is true

- jim_thompson5910

arctan(b/a) = arctan(-1) = ??

- anonymous

-45...

- anonymous

So 315 degrees?

- jim_thompson5910

yes

- anonymous

So would the answer be:
3SqRt2(cos7pi/4 + i sin7pi/4) ?

- jim_thompson5910

correct

- 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

that would be the radian form

- jim_thompson5910

you're welcome

- anonymous

Yes, that is what the answers are listed as.

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