Fresnel Lens project

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Fresnel Lens project

Postby wfws on Wed Jul 20, 2011 2:14 am

I have a big project- worth half a test- and I need some help. I don't want anyone to do the work for me- just help me get unstuck, and point to what I should be looking for.
The project gives a few illustrtions, and then asks some questions. The actual assignment can be found at http://books.google.com/books?id=_cIT_pRdfJIC&pg=PA573&lpg=PA573&dq=Fresnel+lens+trig+cohen&source=bl&ots=HjLGbQWcT9&sig=aPNKGMwAoAsfyvsEX3aCJCW5joE&hl=en&ei=MDgmTrnzO6nmiALe04j0CQ&sa=X&oi=book_result&ct=result&resnum=1&ved=0CBkQ6AEwAA#v=onepage&q&f=false.

On Exercise 1, part B, I get the first part- its from basic geometry that these angles are equal. But the next ones, asking to verify that 02 = B + A2, and so on, I just don't know where to start.
Should I be looking for geometry, or rotating around a unit circle to understand these? Or is this a matter of using some identities?
Sorry I can't ask a better question- I'm just stumped at this point.

I'm sure I"ll have more questions. I'm having trouble knowing how to apply the trig I DO know to this problem.

Thanks for your help!

Wayne
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Postby stapel_eliz on Wed Jul 20, 2011 1:11 pm

Note that is measured from the horizontal and is measured from the vertical. This gives you the location of a right angle.

Label the angle between the terminal rays for and as . Create two equations for this right angle. Subtract, and see what you get. :wink:
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Re: Fresnel Lens project

Postby wfws on Wed Jul 20, 2011 3:44 pm

Thanks so much. I'll give that a try.

I have a test tomorrow, so I'll have to set this aside till after the test. But that's what I was looking for.

Stress level decreasing..

Wayne
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Re: Fresnel Lens project

Postby wfws on Sat Jul 23, 2011 1:34 am

Arggh. If the right angle you identified is created by the initial side of angle Beta, and the intial side of angle theta 2, are the equations you suggest to find angle alpha? Maybe (theta-90) - (beta-90)? to get angle alpha? Then the ref angle of phi2 = alpha = ref angle of (beta + theta2)?

Is that it?
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Postby stapel_eliz on Sat Jul 23, 2011 12:58 pm

wfws wrote:Maybe (theta-90) - (beta-90)? to get angle alpha?

Since and are clearly less than ninety degrees, these subtractions would give you negative values, so I'm not sure where you're going with this...?

Instead, try looking at the two right angles, and creating addition equations which relate to each of these angles. Then see where that leads. :wink:
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Re: Fresnel Lens project

Postby wfws on Sat Jul 23, 2011 3:12 pm

Do you do whiteboard on line tutoring?
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Re: Fresnel Lens project

Postby wfws on Sat Jul 23, 2011 5:32 pm

Here's another try- thanks for the suggestion. So, from the first right angle, I can make the addition equation phi2 + alpha = 90. The second equation would be beta + alpha + theta2 = 90. Is this what you mean by addition equations?
Sorry to be so dense, but I'm not getting how these show how phi2 = beta + theta 2. Can you give a related example?
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Postby stapel_eliz on Sat Jul 23, 2011 7:45 pm

wfws wrote:...from the first right angle, I can make the addition equation phi2 + alpha = 90. The second equation would be beta + alpha + theta2 = 90. Is this what you mean by addition equations?

Yes. Now, since 90 = 90, create an equation with the angles, and solve for the angle you need. You should get one cancellation (from the angle shared by the two additive expressions), and then you should arrive at the desired result. :wink:
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Re: Fresnel Lens project

Postby wfws on Sat Jul 23, 2011 9:43 pm

Thanks I got that one now. Next, I need to show that phi'1 + phi'2 = beta. If the same strategy applies, I can see a right angle formed from the origin of phi'2 and the terminal side of beta. I can see another from the origin of phi'1 to the vertical line that is 90deg. But I can't see how these share anything that will create or result in Beta. What should I be looking for?
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Postby stapel_eliz on Sat Jul 23, 2011 10:59 pm

wfws wrote:What should I be looking for?

You should be looking for stuff you can try. There is no simple "formula"; you have to do some thinking.

At a guess, I'd start with the angle external to that tiny triangle with the two "primed" angles within, and see if I could relate something to the right triangle formed by the horizontal line (extended) below and the right angle formed by the slanty lines above .

Now: What can you come up with? :wink:
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