Question

Let f(x) = sin(x) + icos(x). Show that |f(x)| = 1.

I'm not even sure where to start here...

Answer 1

hey its too much easy... |f(x)|=sqrt((1)^2+(1)^2)=sqrt(2)

Answer 2

here we will take its real part 1 will cos(x) and also 1 from imagnary part i(sinx)
hey u have written wrong equation.. its cos(x)-isin(x)

Answer 3

The question is right. I know DeMoivres, but in this case it's sin(x) + icos(x).

Answer 4

ah oookkk ya its absolute value

Answer 5

So far I have LHS = \[\frac{ e ^{ix}-e^{-ix} }{ 2 } + \frac{ e ^{ix}+e^{-ix} }{ 2 }i\]

Have I completely overthought the problem ? :)

Answer 6

no i is denominator from demorvie`s, but in this case I did`nt get u how u got this result

Answer 7

ok am gonna solve its prove and implement according to this condition the `ll msg u here ok

Answer 8

Thanks :) I'm so confused...

Answer 9

hahahhah i just wanna say write i in denominator ok i mean (eix−e−ix)2+(eix+e−ix)2i

Answer 10

Hmm... I don't get it. How does that show that |sin(x) + icos(x)| = 1?

Answer 11

oh i see wait i solve it and post the diagrame okk buddy

Answer 12

Anyone?

Answer 13

daer i can`t .. i have seen the same result for it sqareroot 2 iz answer for both condition.. cos when we see cosx+isinx,, or sinx+icosx... taking coifficients we get the same result..

Answer 14

i thinks its wrong question or some thing missing ... can u give the link where did u got it from ???

Answer 15

Yep, it's question 1 iv) from this paper: http://www.scribd.com/doc/112148611/MATH1131-Mathematics-1A-1 Thanks :)

Answer 16

pdf attached here.

Answer 17

http://www.mathwarehouse.com/ellipse/equation-of-ellipse.php
http://www.algebra.com/algebra/homework/equations/THEO-20100329.lesson

Answer 18

u got ur answer or not /??

Answer 19

Nope, heh. Not yet :(

Answer 20

ok if its difficault then msg me ur question i will solve it my self and will post the pic here
ok

Answer 21

Yeah I don't know how to get the function to = 1...

Answer 22

I am thinking that you are using the fact that |f(x)| represents the exact distance between it and 0 on the complex plane...
Notice that you can create a right triangle with the real part of f(x) and the imaginary part of x on a complex plane:

So, here, Re(f) = sin(x) and Im(f) = cos(x). We drop off the i since we're dealing with distances only for the imaginary part.
Using Pythagorean theorem, |f(x)|^2 = (Re(f))^2 + (Im(f))^2 = sin^2 x + cos^2 x. Since we have a neat trigonometric identity here, we can use it to reduce sin^2 x + cos^2 x as 1. Take the square root of both sides and we're done. :)