Question

rcos((theta)-pi/6)=2???? convert to rectangular?

Answer 1

multiply the complex no z = x+iy by e^(-ipi/6)

Answer 2

\[rcos (\Theta-\pi/6)=2\]

Answer 3

cos(t-pi/6)=2

Answer 4

you can always remember cos(a+b) = cosa * cosb - sina * sinb

Answer 5

let the new complex no be z1
the x part of z1 is rcos(theta - pi/6)

Answer 6

its just a straight line right?

Answer 7

no clue:(

Answer 8

@franc..wat happnd 2 the r in ur prev step?

Answer 9

since cos only goes from -1 to 1; i think its a false statement

Answer 10

wait

Answer 11

you can expand the cosine of sum of angles...then you would had something like ax+by = 2

Answer 12

r is sqrt(x^2 + y^2)
cos (t - pi/6) is ((root 3)x + y) / 2

Answer 13

cos(doesnt matter) never = 2

Answer 14

.did u get wht i wrote?

Answer 15

remember r could be 2, then let theta = pi/6 ...done 2 = 2

Answer 16

soo..in rectangular form..tht is....?

Answer 17

whom r u asking?

Answer 18

doesn't matter, the fact is you need to know what you have to do....have any idea?

Answer 19

i told u guys

Answer 20

this is not a site where you get answers, we discuss possible ways to find out the solution

Answer 21

just rotate a complex no by pi/6 clockwise to get cos(theta - pi/6)

Answer 22

ohhok..\[r=2/((\sqrt{3}/2)\cos \Theta-1/2(\sin \Theta))\]

Answer 23

well, you got something (: now remember that r cos(theta) = x and r sin (theta) = y

Answer 24

there is no more thinking in this problem

Answer 25

z = r(cost + isint)
let x = rcost
y = rsint
now
ze^(-ipi/6) = r[cos(t-pi/6) + isin(t - pi/6)]

Answer 26

if z= x+ iy
\[ze^{-i \pi/6} = [\sqrt{3}x + y]/2 + i [\sqrt{3}y - x]/2\]

Answer 27

getting it?

Answer 28

somewhat...is der a way to do it without the i's?

Answer 29

none tht i know

Answer 30

so there u go
if u compare the two eqns
u find
\[rcos(\theta- \pi /6) = [\sqrt{3}x + y]/2\]

Answer 31

thats equal to 2

Answer 32

ohhok..yeah i think i got it! thanks a lott

Answer 33

√3x + y = 4

Answer 34

no problem...just be comfy with complex nos and ull be fine