Trigonometry problem

by Pranav-Arora
Tags: euler's formula
P: 3,806
 Quote by tiny-tim Because the question defines n as being the highest value of m. The "highest" one is C6 (any C above C6 is zero), so n = 6.
Okay, but if i wanted to find out the value of C1?
 Sci Advisor HW Helper Thanks P: 26,157 C1 is the coefficient of cosx, so it's 0.
P: 3,806
 Quote by tiny-tim Because the question defines n as being the highest value of m. The "highest" one is C6 (any C above C6 is zero), so n = 6.
Ok thanks..
You helped me a lot!!
Thank you tiny-tim!!
 Sci Advisor HW Helper Thanks P: 26,157 As an "appendix", here's the alternative result using complex numbers and Euler's formula … you won't understand it yet, but anyone else following this might be interested … sin3xsin3x = sin3x Im(e3ix) = sin3x Im((cosx + isinx)3) = Im((cosxsinx + isin2x)3) = (1/8) Im((sin2x + i(1 - cos2x))3) = (1/8) Re((isin2x + 1 - cos2x)3) = (1/8) Re((1 - e2ix)3) = (1/8) Re(1 - 3e2ix + 3e4ix - e6ix) = (1/8) (1 - 3cos2x + 3cos4x - cos6x) This works for sinkxsinkx for any value of k … sinkxsinkx = (-1/2k) ∑ kCm cos2mx (so what's sinkxcoskx, and is there a quicker way of doing it?)
P: 3,806
 Quote by tiny-tim As an "appendix", here's the alternative result using complex numbers and Euler's formula … you won't understand it yet, but anyone else following this might be interested … sin3xsin3x = sin3x Im(e3ix) = sin3x Im((cosx + isinx)3) = Im((cosxsinx + isin2x)3) = (1/8) Im((sin2x + i(1 - cos2x))3) = (1/8) Re((isin2x + 1 - cos2x)3) = (1/8) Re((1 - e2ix)3) = (1/8) Re(1 - 3e2ix + 3e4ix - e6ix) = (1/8) (1 - 3cos2x + 3cos4x - cos6x) This works for sinkxsinkx for any value of k … sinkxsinkx = (-1/2k) ∑ kCm cos2mx (so what's sinkxcoskx, and is there a quicker way of doing it?)
I think i would take a look at it again. My teacher may start up with complex numbers after 2 or 3 chapters. I would save this a text file in my computer.
 HW Helper P: 6,187 As another aside, this series is called the Fourier cosine series expansion. Fourier was a mathematician who found that any function can be expressed as a cosine series of this type, and that the coefficients are unique. Many mathematical programmatic tools have this built in, such as Mathematica. It's also available in WolframAlpha: http://www.wolframalpha.com/input/?i...s+sin^3x+sin3x although I found that you won't be able to get the full expansion here, because WolframAlpha will time out before you'll get there. :( However, I'm afraid it will be quite a while before you get to learn this, and maybe you will never learn this, depending on what you're going to study exactly. The application of Fourier analysis is typically in spectral analysis, signal processing, image processing and electronical engineering, where it's used for frequency related analysis and operations.
 HW Helper Thanks P: 10,377 Hi, tiny-tim, is not there a sign error in your derivation? Just to show off my way: $$sinx=\frac{e^{ix}-e^{-ix}}{2i}$$, $$cosx=\frac{e^{ix}+e^{-ix}}{i}$$, $$sin(x)^3 sin(3x)=(\frac {e^{ix}-e^{-ix}}{2i})^3 \frac{e^{3ix}-e^{-3ix}}{2i}=$$ $$=\frac {e^{3ix}-3 e^{ix}+3 e^{-ix}-e^{-3ix}}{-8i}\frac{e^{3ix}-e^{-3ix}}{2i}=$$ $$1/16(e^{6ix}-3 e^{4ix}+3 e^{2ix}-1-1+3 e^{-2ix}-3 e^{-4ix}+e^{-6ix})=$$ $$1/8(cos(6x)-3cos(4x)+3cos(2x)-1)$$ ehild
HW Helper
P: 6,187
@tiny-tim: Nice improvement on the calculation btw. It's way less work!

 Quote by ehild Just to show off my way:
Thank you ehild, this is a nice solution, and shows a couple of things.

First, I think that Pranav-Arora is very well able to follow this one and do it himself.
It only requires a basic grasp on complex numbers, and beyond that it's just exponentiation rules, which Pranav-Arora's has already practiced quite a bit.

And it also shows that in my humble opinion calculating with sines, cosines and angles is kind of obsolete.
I found that calculating with imaginary e-powers (or vectors in geometric problems) is much easier.

I'm interested in examples where you would really want to use sines, cosines and angles to do your calculations....
Does any of you have any?
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P: 10,377
 Quote by I like Serena And it also shows that in my humble opinion calculating with sines, cosines and angles is kind of obsolete. I found that calculating with imaginary e-powers (or vectors in geometric problems) is much easier. I'm interested in examples where you would really want to use sines, cosines and angles to do your calculations.... Does any of you have any?
What about determining the height of a tower from the angle of view at a known distance?

ehild
HW Helper
P: 6,187
 Quote by ehild What about determining the height of a tower from the angle of view at a known distance? ehild
Yes, it's still useful for immediate applications of the definitions of sines, cosines and tangents based on an angle.
What I'm looking for is actually calculating with it.
Usually I convert an angle as soon as I can to a vector, then calculate with it, and possibly convert it back into an angle to present a result.
To me it feels a bit like converting inches to metric before calculating, and converting back to present a result.

And even here...
If you measure an angle, you use a tangent to convert distance to height.
But what you actually measure, is a distance on a measuring device that is marked as an angle, but it might as well be marked as a distance at a specific short distance, or as a tangent ratio.
Multiply by the ratio in distance and you get the height without ever using an angle or the tangent function.
 Sci Advisor HW Helper Thanks P: 26,157 hi ehild! yes that's nice and it's certainly shorter for k = 3 (but not so good for large k)
 P: 436 Please try to use euler's forumla and u will realise that it is made of some even m (2,4,6). So n=6
HW Helper
P: 6,187
 Quote by tiny-tim and it's certainly shorter for k = 3 (but not so good for large k)
Let's see.

For k is odd we have:
$$\begin{eqnarray} \sin^kx \sin kx &=& \left(\frac {e^{ix} - e^{-ix}} {2i} \right)^k \left(\frac {e^{ikx} - e^{-ikx}} {2i} \right) \\ &=& \frac 1 {(2i)^{k+1}} \left(\sum_{m=0}^k \binom k m (-1)^m e^{i(k-2m)x} \right) (e^{ikx} - e^{-ikx}) \\ &=& \frac 1 {(2i)^{k+1}} \sum_{m=0}^k \binom k m (-1)^m (e^{i(2k-2m)x} + e^{-i(2k-2m)x}) \\ &=& \frac 2 {(2i)^{k+1}} \sum_{m=0}^k \binom k m (-1)^m \cos(2k-2m)x \\ &=& (-1)^{\frac {k+1} 2} \frac 1 {2^k} \left(\binom k 0 \cos 2kx - \binom k 1 \cos(2k-2)x + \binom k 2 \cos(2k-4)x - ... + \binom k {k-1} \cos 2x - 1 \right) \end{eqnarray}$$

As you can see, I've also corrected the sign error.
HW Helper
Thanks
P: 26,157
mmm … let's compare length …

sinkxsinkx = sinkx Im((cosx + isinx)k)

= Im((cosxsinx + isin2x)k)

= (1/2k) Im((±sin2x + i(1 - cos2x))k)

= (1/2k) Im(ik(1 - cos2x ± isin2x)k)

(I lost an i2 here originally )

= (1/2k) Im(ik(1 - e2ix)k)

= (1/2k) Im(ikkCm (-1m) e2imx)
= (1/2k)(-1)(k-1)/2kCm (-1m) cos2mx for k odd

= (1/2k)(-1)k/2kCm (-1m) sin2mx for k even
ooh, yes, yours is 2 lines shorter
 Quote by I like Serena $$\frac 1 {(2i)^{k+1}} \sum_{m=0}^k \binom k m (-1)^m (e^{i(2k-2m)x} + e^{-i(2k-2m)x})$$ $$\frac 2 {(2i)^{k+1}} \sum_{m=0}^k \binom k m (-1)^m \cos(2k-2m)x$$
oh, that's how it works! nice!

(and for k even, there's a sign-change somewhere)

though we can then use kCm = kCk-m to go straight to …

$$\frac 2 {(2i)^{k+1}} \sum_{m=0}^k \binom k m (-1)^{k-m} \cos2mx$$
 HW Helper Thanks P: 10,377 I just wanted to show this, but you both beat me while I watched Poirot on TV ehild
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P: 10,377
 Quote by I like Serena If you measure an angle, you use a tangent to convert distance to height. But what you actually measure, is a distance on a measuring device that is marked as an angle, but it might as well be marked as a distance at a specific short distance, or as a tangent ratio..
As far as I know the angle of view is measure of the length of an arc between the arms of a compass, when those arms point to the desired directions. And the tangent of that angle is used to get distance or height.

ehild
 P: 3,806 Hi!! I had read the complex numbers chapter from my textbook and i think i now have a basic idea of them. So may i know how ehild has got this relationship:- $$sinx=\frac{e^{ix}-e^{-ix}}{2i}$$
 HW Helper P: 6,187 Hi again! Did you find Euler's formula eix = cos x + i sin x in your chapter?

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