Question

derive sin(x+jy)

Answer 1

do you mean derive the derivative?

Answer 2

i.e. differentiate?

Answer 3

or do you mean show how we derive the function sin(x+jy) from triangles?

Answer 4

derive as in derive that function i don't know how to start pls help :) @zzr0ck3r

Answer 5

I would google it. It takes allot of pictures

Answer 6

You can first derive the formula for sin(A+B).
Then put A = x and B = jy to derive sin(x+jy).

Here is one link that derives sin(A+B): http://www-istp.gsfc.nasa.gov/stargaze/Strig5.htm

Answer 7

sin(A+B), then? im sorry i still have no idea .. the link you gave confuses me :((

Answer 8

can i use this as my solution? @aum http://www.symbolab.com/solver/derivative-calculator/%5Cfrac%7Bd%7D%7Bdx%7D(sin(x%2Bjy))/?origin=button

Answer 9

@KarlaKalurky is this a calculus or trigonometry course?

Answer 10

advance mathematics

Answer 11

@Kainui

Answer 12

@amistre64

Answer 13

@ganeshie8

Answer 14

derive as in "differentiate" ?

Answer 15

yes

Answer 16

you have 3 variables in sin(x+jy)

Answer 17

can you take a screenshot of original question and attach if psble ?

Answer 18

okay :))) wait

Answer 19

waiting

Answer 20

my sr codes ends with an even no. so I must derive for (x+jy) :))))

Answer 21

Derive doesn't mean differentiate. I don't know why people keep associating them.

Answer 22

how ? where shall i start? pleasee help me :(

Answer 23

Do you know what \(j\) is?

Answer 24

complex number, imaginary number..
i

Answer 25

Okay, so what concepts have they introduced so far?

Answer 26

Can you use the Maclaurin series?

Answer 27

converting complex nos. to rectangular, polar, trigo, and expo form... ??
i think so :)

Answer 28

hey i think they just want you derive the formula
they are not asking you to find derivATIVE

Answer 29

how?? :)

Answer 30

This is sort of the definition for a real number \(t\).

Answer 31

j is the imaginary unit ?

Answer 32

So how do you think we would define it for a complex number then? Would we have a unit sphere?

Answer 33

Okay, maybe we can consider how...

Answer 34

\[
x+jy = re^{j\theta} = \cos(\theta)+i\sin(\theta)
\]

Answer 35

However, if we let \(\theta\) be an imaginary number...

Answer 36

Maybe something like...\[
re^{j(x+jy)} = \cos(x+jy)+j\sin(x+jy)
\]

Answer 37

\[
re^{j(x+jy)} = re^{xj-y} = (re^{-y})e^{jx}
\]

Answer 38

@KarlaKalurky Does this help?

Answer 39

ahh why did you substitute to re^j(theta)?

Answer 40

Hmm, well

Answer 41

Actually, I made a lot of mistakes, can I just start over real quick?

Answer 42

super duper fine! :DD

Answer 43

We start with this:\[
e^{j\theta} = \cos(\theta)+j\sin(\theta)
\]

Answer 44

We take the imaginary part of both sides: \[
\Im(e^{j\theta}) = \sin(\theta)
\]

Answer 45

Follow so far?

Answer 46

why not simply use the sin(A+B) formula ?

Answer 47

ok..

Answer 48

Now we need to use \(\theta=x+jy\):\[
\Im(e^{j(x+jy)})=\sin(x+jy)
\]So, now we have found the answer.

Answer 49

We can simplify a bit though...

Answer 50

\[\sin(x+jy) = \sin x \cos(jy) + \cos x\sin(jy) = \sin x \cosh y + i\cos x \sinh y \]

Answer 51

Yeah sure, let's just assume that those properties exist.

Answer 52

It's not like we need to prove it or anything.

Answer 53

hyperbolics? @ganeshie8 Nice!

Answer 54

okay.. i think i saw something like that in our manual that sin

Answer 55

*that sinh and cosh

Answer 56

you should know their definitions since you're doing advanced mathematics :P

Answer 57

usually mathematicians use i form imaginary unit, are you doing engineering ?