Question

@DLS

Answer 1

arre aagayaa

Answer 2

message nahi ja raha tha :/

Answer 3

\[\LARGE \cos( \frac{-\pi}{2})+ \sin (\frac{-\pi}{2})\]

Answer 4

cos(-x)=cosx

Answer 5

You have \(a + bi\) and the polar form as \(r(\cos\theta + i\sin\theta)\).

First find \(|a + bi|\) which is \(r\). Then express the complex number as \(r\left(\dfrac{a}{r} + \dfrac{bi}{r}\right)\). Then you can find \(\theta\) which is \(\arctan \left(\dfrac{b}{a}\right)\).

Answer 6

:-O

0 in the denominator

Answer 7

I got that answer by hit and trial. :-\

Answer 8

maine sahi to kiya anyway

Answer 9

BTW, your answer must not be in the form \(r(\cos \theta - i\sin \theta)\)

Answer 10

-pi 2 ya 3pi/2

Answer 11

If they want \(0 \le \theta \le 2\pi\) then 3pi/2.

Answer 12

they didnt ask anything likethat

Answer 13

both are correct right?

Answer 14

Then write \(\dfrac{3\pi}{2}\). Negatives are commonly not done in polar forms. :-\

Answer 15

(Y)

Answer 16

I just chose a \(\theta\) such that \(\cos \theta = 0\) and \(\sin\theta = -1\).

Answer 17

Sab theek?

Answer 18

yeah,gotcha