Question

Trigonometrical forms: -

Answer 1

@terenzreignz sir I learned how to convert a complex number into a polar or trigonometrical form.
But now I have a problem.
would u solve; plz?

Answer 2

from u.

Answer 3

Would you like to show me?

Answer 4

^_^
\[\sin 32^o+ i \cos 32^o\]

Answer 5

plz tell me how to convert it into trigo form?

Answer 6

Trigonometric form is...
\[re^{i \theta}\]
Is that what you meant?

Answer 7

no \[\cos \theta + i \sin \theta\]
this one is which i want:)

Answer 8

For a 'normal' complex number of
a + bi, how do you normally convert it to trigonometric? :)

Answer 9

\[\sqrt{a^2+b^2}[\cos (\tan ^{-1}\frac{b}{a})+ i \sin (\tan^{-1}\frac{b}{a})]\]

Answer 10

well, no matter how complicated it looks, I suggest you do that with your
sin 32 and cos 32 as well :)

Answer 11

Oh, and feel free to ask me further questions if you get stuck :)

Answer 12

but one my friend did like that
sin 32 =cos 58
cos 32 =sin 58
so
question will look like
cos 58 + i sin 58
which in fact is answer which is given in my book
but i wanna know
why did he converted them like that?

Answer 13

Oh ok, that works (and is much faster than what I did, I'm sorry )

Answer 14

np:)

Answer 15

That's because by converting, you've suddenly turned it into the form you desired
that is to say
cos theta + isin theta

Answer 16

no no i don't want that
I want why he wrote cos 32=sin 58
& sin 32 = cos 58

Answer 17

ahh ok:)
You know that
\[\sin(A + B) = sinA cosB + cosA sinB\]
right?

Answer 18

yeah:)

Answer 19

Well
\[\sin(90^{o} - A) = \sin90^{o}cosA - \cos90^{o}sinA\]
cos 90 = 0
sin 90 = 1, so
\[\sin(90^{o} - A) = cosA\]

Answer 20

Do you understand now? :)

Answer 21

k! i gt it now.
thanx a lot:)

Answer 22

No problem :)