Solving a Trigonometry Question: Understanding the Equation and Breaking it Down

[yungman]

From the equation, it implies:
\left[\frac{\cos\left(\frac{\pi}{2}\cos\theta\right)}{\sin\theta}\right]^2=\sin^3\theta
For the life of me, I have no idea how this come about.

I don't even know where to start, this implies $\cos\left(\frac{\pi}{2}\cos\theta\right)=\sin^{\frac 5 2}\theta$.

Please help me in how to even start. How do you break up $\cos\left(\frac{\pi}{2}\cos\theta\right)$?

Thanks

 

[Curious3141]

From the equation, it implies:
\left[\frac{\cos\left(\frac{\pi}{2}\cos\theta\right)}{\sin\theta}\right]^2=\sin^3\theta
For the life of me, I have no idea how this come about.

I don't even know where to start, this implies $\cos\left(\frac{\pi}{2}\cos\theta\right)=\sin^{\frac 5 2}\theta$.

Please help me in how to even start. How do you break up $\cos\left(\frac{\pi}{2}\cos\theta\right)$?

Thanks

For a start, it's not even true (in general), i.e. it's not an identity.

Try $\theta = 23$ (radians), for example. The LHS is positive, the RHS is negative, but the numerical values are different as well.

Sure this is not an equation you have to solve?

Even then, an exact solution is not possible, and you'll have to use approximate numerical methods.

 

[yungman]

For a start, it's not even true (in general), i.e. it's not an identity.

Try $\theta = 23$ (radians), for example. The LHS is positive, the RHS is negative, but the numerical values are different as well.

Sure this is not an equation you have to solve?

Even then, an exact solution is not possible, and you'll have to use approximate numerical methods.

Thanks for your reply, I don't even know how to use numerical method as it is a cosine of a cosine.
This is the copy of the Antenna Theory by Balanis page 162. This is a very famous book, pretty much one of the two books for Antenna.

 

[Mark44]

Your attachment is a bit grainy, but it looks like they are saying that the two expressions are "approximately equal" (≈) rather than exactly equal (=).

Is there some information about θ given that you didn't show in the attachment? If so, that would be helpful.

 

[yungman]

Your attachment is a bit grainy, but it looks like they are saying that the two expressions are "approximately equal" (≈) rather than exactly equal (=).

Is there some information about θ given that you didn't show in the attachment? If so, that would be helpful.

Thanks.

$\theta$ is from 0 to π. This is a full spherical coordinates. This is about EM wave radiating from the origin. So there is no more restriction on $\theta$.

 

[Mark44]

My best guess is that they are using two or three terms in the Maclaurin series for cos($\pi/2$~cos(θ)), like so:

$cos(\pi/2~cos(θ)) = 1 - (\pi/2)^2/2! * cos^2(θ) + (\pi/2)^4/4! *cos^4(θ) -+ ...$

 

[yungman]

This is a copy that contains a little more information. I have some hand written notes to show where the equations come from. Basically $W_{av}$ is the average energy density at direction given by $\theta$.

W_{av}(\theta)=\frac 1 2 Re[\vec E_{\theta}X\vec H_{\phi}^*]=\hat R\frac{|E_{\theta}|^2}{2η}

For $\frac {\lambda}{2}$ dipole, $kl=\frac{\pi}{2}$ that give the $W_{av}$ formula.

I don't know all these help at all as I just extract the part in question. I just don't see how more explanation help. But I post this just in case I missed something important. Again thanks for all the help. I know this is not an easy problem. These are advanced electromagnetics in grad school...and I am rusty in math.

 

[yungman]

My best guess is that they are using two or three terms in the Maclaurin series for cos($\pi/2$~cos(θ)), like so:

$cos(\pi/2~cos(θ)) = 1 - (\pi/2)^2/2! * cos^2(θ) + (\pi/2)^4/4! *cos^4(θ) -+ ...$

Thanks, still this will not give the answer from the book!

 

[Dick]

Thanks, still this will not give the answer from the book!

I really don't think that relation is 'derived' at all. The left hand side of your expression approaches 0 at θ=0, it's 1 at θ=pi/2 and it approaches 0 again at θ=pi. So it's bound to look something like a sine. If you plot it together with (sinθ)^2 you'll see that it's too broad around θ=pi/2, (sinθ)^4 is too narrow. (sinθ)^3 happens to fit pretty well. It's not so good at the ends, but ok.

It's just a convenient approximation, because it's pretty hard to deal with things like cos(cos).

 

[yungman]

I really don't think that relation is 'derived' at all. The left hand side of your expression approaches 0 at θ=0, it's 1 at θ=pi/2 and it approaches 0 again at θ=pi. So it's bound to look something like a sine. If you plot it together with (sinθ)^2 you'll see that it's too broad around θ=pi/2, (sinθ)^4 is too narrow. (sinθ)^3 happens to fit pretty well. It's not so good at the ends, but ok.

It's just a convenient approximation, because it's pretty hard to deal with things like cos(cos).

Thanks for the detail explanation. What program do you use to plot this, its there any free program on line that I can use?

 

[Dick]

Thanks for the detail explanation. What program do you use to plot this, its there any free program on line that I can use?

Lots, I would imagine. I use Maxima for stuff like this. But you do have to install it on your computer, I don't think it's online. It's VERY free (not just in the sense of no charge, but it's also 'open source'). It also does computer algebra and an endless number of other things. I've used it ever since people stopped paying for a copy of Mathematica for me. Here's a plot command:

plot2d([sin(t)^3,(cos(%pi/2*cos(t))/sin(t))^2],[t,0.01,3]);

I'm sure other people have their favorites as well. But you should definitely check it out. Would be pretty handy for antenna design.

WAIT. I was wrong. Here's an online version of Maxima. http://maxima-online.org. Just paste the line I gave you into it and play around with the exponent '3'.

 

[yungman]

Thank you so very much, I'll learn this and see what happen. Also I see the "≈" instead of "=".

Alan

 

[Dick]

Thank you so very much, I'll learn this and see what happen.

Alan

Very welcome! There is a learning curve, but it's worth the time. You definitely need this.