Math Problem

Given the values provided by Excel, can't you try using ratio and proportion?

Yes, I think that will work.

I was just looking for a "proper" way to calculate it directly. However, that may not be possible with the sine and cosine functions having a ratio between two angles in one formula.

Sounds like a homework problem, so I am just going to sketch the analytical, closed form solution.

You can use a basic trig formula for cos (a+b) and another basic trig formula for cos (2a) and sin(2a) to get from cos (3a) to an expression that is a function of only cos (a). Once you have that, you can solve for cos(a) and you will find the answer is 60 degrees.

Thanks, but no, it is not home work.

I am long out of school and this is a partial trochoidal shape.

ok. then: cos (A + B) = cos A cos B - sin A sin B.

assuming A to be 2 B (to get to my 3 alp) we get:

(cos 2B cos B - sin 2B sin B) divided by (cos B) I get:

(cos 2B - sin 2B tan B) = -2. Then cos (2B) = 1 - 2 sin^2 2B

so: 1 - 2 sin^2 2B - sin 2B tan B = - 2. Now what?

By the way, I was wondering where USSA is, perhaps that? However, no offence implied.

http://www.amazon.ca/Back-USSA-Eugene-Byrne/dp/0929480848

The next step from 1 - 2sin^2 2B - sin 2B tan B = -2 is:

2 sin^2 2B + sin 2B tan B = 3.

How do we get angle B from that?

emc c, thanks you very kindly for that.

However, when checking on Excel it is only correct at that point of 60 degrees and -2 of the original question. Using as answer -2.2 or -1.9 does not provide the original angles that produce these results.

A GA vote nevertheless.

At one time I thought it might be a quadratic equation of sin when looking at my last deduction of: 2 sin^2 2B + sin 2B tan B = 3.

But that "tan B" factor is in the way.

Regards

Dear emc c

I am sory, but I have to apologize. I made a mistake in the intrepertation of your solution. It works out fine, as is well. Sorry for being too hasty.

emc c, Thanks for taking the time to elucidate. I am not an engineer or a math major but I enjoy math. I like trying to solve the problems posted here but often lack the knowledge. Often I see answers that say " here this is how easy it is" but don't explain the process. I appreciate your time and efforts.

MusicAl

There is more than one answer to this problem.

Take the last step of this solution:

cos2 (a) = 1/4

from which follows

cos (a) = 1/2, from which the answer follows a = 60 degrees

more correctly this should be

from which follows

cos (a) = ±1/2, from which the answers follow a = +60, -60 degrees

Furthermore since cos(a)=cos(a + N*360°) the complete answer is

a = ±60° ± N*360° where N is any integer

Always be cautious when square roots appear; it isn't always evident which root is the one you need (or both may be valid)

Strictly speaking you are of course correct. There are two answers.

Since it was related to a practical question I was happy with the answer as provided by emc c, specifically the sequence of steps that give the answer.

Going above 90 degrees results in the "mirror" value. E.g. 95 degrees = 85 degrees, 100 degrees = 80 degrees etc.

Regards

I dont know who voted you, but if I were teacher, I would have given you only 69 scrore, as you forgot the angle is periodic, at least you would add a nx 2pi,

where n=+-1,+-2,+-3...

its really a high school homwwork,

cnpower, it looks like you get an even worse grade, because you didn't read the prior posts. Sliderule already made that point, but more to the point, emc-c didn't provide the complete derivation until he was assured it wasn't a homework problem; but a real one instead.

His first answer outlines the solution with plenty of information for the student to fill in the missing steps. After being told a real problem was being solved, he supplied the complete derivation, and at that point only the fundamental answer was of interest, not all the harmonics.

And the original poster verified that was indeed what he was after.

very sorry for my thread,

your critique is right. I really havnt read the following threads. Slideruler finished right answer, as well as emcc.

I know emc-c is serious for every issue. but I like to make a joke, he may know this and would not care about.

In MathCAD it would be

Your "icons" are all dead. Re-post?

you can solve alp with this equation: alp= "arccos(return)/4 " but this only works if the return is -1.. other means of solving alp is by using a scientific calculator..