Question

Write this in polar{trigonometrical form}: -


\[-\sin 110^o+i \cos 110^o.\]

Answer 1

evaluate is what you need to do

Answer 2

but i have done many questions
but this is different

Answer 3

actually, it is in polar form already

Answer 4

weird that it says "write in trig from" when it is already in trig form

Answer 5

no; i tell u its answer but i don't know how can i gt that

Answer 6

hold the phone.
it says "write in trig form" and i am looking at trig functions right?

Answer 7

\[\cos (360^ok+200^o)+i \sin (360^ok+200^o).\]

Answer 8

you could rewrite in standard from as \(a+bi\) or you could write as \(e^{i\theta}\) if you like

Answer 9

ooooooooooooooooh i see !

Answer 10

sorry
my fault.
they want you to write this as \(\cos(\theta)+i\sin(\theta)\)

Answer 11

np:)
but now how to get it?

Answer 12

where the hell this 200 came from?

Answer 13

ok lets see
\(\cos(x)=\sin(90-x)\) so \(\cos(110)=\sin(90-110)=\sin(-20)\) no that doesn't do it, but it is a start

Answer 14

k

Answer 15

that is the idea though, switch from cosine to sine and vice versa

Answer 16

ya u r right

Answer 17

you got it from here?

Answer 18

no

Answer 19

dumbass question if you don't mind me saying so

Answer 20

lol

Answer 21

weird that i am getting a slightly different answer

Answer 22

Oh i gt it:)
\[-\sin 110^o+i \cos 110^o.\]\[\implies - \cos (90-110)+i \sin (90-110)\]\[\implies - \cos (-20) + i \sin (-20)\]\[\implies \cos (180 - (-20))+ i \sin (180-(-20))= \cos (360^ok+200^o)+i \sin (360^ok+200^o)\]

Answer 23

\(-\sin(110)=\sin(-110)\) because sine is odd. to switch to cosine we can use \(\cos(90-x)=\sin(x)\) so \(90+110=200\) ok got it

Answer 24

oh and you got it too
faster than i did

Answer 25

still a dumbass question, has nothing at all to do with complex numbers, just changing cosines to sines etc

Answer 26

lol u helped me lot.
thanx a lot:)