Trigonometry problem

tiny-tim

I just read it off your answer of (1/8)cos6x -(3/8)cos4x +(3/8)cos2x - (1/8)

Pranav-Arora

Okay, i understand how you get C₆=(1/8).
But how do you get n=6??

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.

Pranav-Arora

Okay, but if i wanted to find out the value of C₁?

tiny-tim

C₁ is the coefficient of cosx, so it's 0.

Pranav-Arora

Ok thanks..
You helped me a lot!!
Thank you tiny-tim!!

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(e^3ix)

= 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 - e^2ix)³)

= (1/8) Re(1 - 3e^2ix + 3e^4ix - e^6ix)

= (1/8) (1 - 3cos2x + 3cos4x - cos6x)

This works for sin^kxsinkx for any value of k …


(so what's sin^kxcoskx, and is there a quicker way of doing it?)

Pranav-Arora

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.

I like Serena

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.

ehild

Hi, tiny-tim, is not there a sign error in your derivation?

Just to show off my way:

[tex]sinx=\frac{e^{ix}-e^{-ix}}{2i}[/tex],
[tex]cosx=\frac{e^{ix}+e^{-ix}}{i}[/tex],

[tex] sin(x)^3 sin(3x)=(\frac {e^{ix}-e^{-ix}}{2i})^3 \frac{e^{3ix}-e^{-3ix}}{2i}=[/tex]

[tex]=\frac {e^{3ix}-3 e^{ix}+3 e^{-ix}-e^{-3ix}}{-8i}\frac{e^{3ix}-e^{-3ix}}{2i}=[/tex]
[tex]1/16(e^{6ix}-3 e^{4ix}+3 e^{2ix}-1-1+3 e^{-2ix}-3 e^{-4ix}+e^{-6ix})=[/tex]
[tex]1/8(cos(6x)-3cos(4x)+3cos(2x)-1)[/tex]

ehild

I like Serena

@tiny-tim: Nice improvement on the calculation btw. It's way less work!

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?

ehild

What about determining the height of a tower from the angle of view at a known distance?

ehild

I like Serena

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.

tiny-tim

hi ehild!

yes that's nice

and it's certainly shorter for k = 3 (but not so good for large k)

icystrike

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

I like Serena

Let's see.

For k is odd we have:
[tex]\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}[/tex]

As you can see, I've also corrected the sign error.

tiny-tim

mmm … let's compare length …

sin^kxsinkx = sin^kx Im((cosx + isinx)^k)

= Im((cosxsinx + isin²x)^k)

= (1/2^k) Im((±sin2x + i(1 - cos2x))^k)

= (1/2^k) Im(i^k(1 - cos2x ± isin2x)^k)

(I lost an i² here originally )

= (1/2^k) Im(i^k(1 - e^2ix)^k)

= (1/2^k) Im(i^k ∑ ^kC_m (-1^m) e^2imx)

= (1/2^k)(-1)^(k-1)/2 ∑ ^kC_m (-1^m) cos2mx for k odd

= (1/2^k)(-1)^k/2 ∑ ^kC_m (-1^m) sin2mx for k even

ooh, yes, yours is 2 lines shorter
oh, that's how it works! nice!

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

though we can then use ^kC_m = ^kC_k-m to go straight to …

[tex]\frac 2 {(2i)^{k+1}} \sum_{m=0}^k \binom k m (-1)^{k-m} \cos2mx[/tex]