Let f(x) = sin(x) + icos(x). Show that |f(x)| = 1.
I'm not even sure where to start here...

- anonymous

Let f(x) = sin(x) + icos(x). Show that |f(x)| = 1.
I'm not even sure where to start here...

- jamiebookeater

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this

and **thousands** of other questions

- anonymous

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

- anonymous

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)

- anonymous

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

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- anonymous

ah oookkk ya its absolute value

- anonymous

So far I have LHS = \[\frac{ e ^{ix}-e^{-ix} }{ 2 } + \frac{ e ^{ix}+e^{-ix} }{ 2 }i\]
Have I completely overthought the problem ? :)

- anonymous

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

- anonymous

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

- anonymous

Thanks :) I'm so confused...

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

Anyone?

- anonymous

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..

- anonymous

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

- anonymous

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

- anonymous

pdf attached here.

##### 1 Attachment

- anonymous

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

- anonymous

u got ur answer or not /??

- anonymous

Nope, heh. Not yet :(

- anonymous

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

- anonymous

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

- AccessDenied

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:
|dw:1352104120632:dw|
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. :)

- anonymous

Thanks so much for your clear and concise answer!! I've been stuck on this question for a long time :)

- AccessDenied

You're welcome! :)

- sirm3d

perhaps this definition can refresh you \[\huge \left| a + bi \right|=a^2+b^2 \]

Looking for something else?

Not the answer you are looking for? Search for more explanations.