Splendid! It makes sense. Thanks, slipstream!

More Trig... sorry for hammering these same identities, but I hear it's really important for calc and I plan to study Applied Math so I need a really good handle on these...

We sometimes get problems that say, simplify this Trig Function: blah... and basically, we "know" only 2 right triangles at this point. One with angles pi/4 & pi/4 and one with angles pi/3 & pi/6. So our approach is to evaluate the Trig Function to see if we recognize a sum or difference using multiples or any of pi/3, pi/4, pi/6. I wish there were a more systematic way to do this, but afaik, there isn't. You just look for it, like factoring a polynomial without the quadratic equation.

For example,

Evaluate: sin(pi/12)
We convert it to sin(pi/3 - pi/4) and plug & chug using the sine difference identity.

Another example using half angles, Evaluate: sin(pi/8)
We convert it to sin[(pi/4)/2] and plug & chug using the sine half angle identity.

Now on to the problem driving me a little nuts...

Evaluate Trig Function: sin(11pi/8)

So, my thoughts here... and some questions...

The 8 in the numerator definitely implies I'm looking for some multiple of pi/4.

I can do this: sin(11pi/8) = sin(pi + 3pi/8)

Now, if I had sin(3pi/8) in isolation, I could use the half angle formula on this pretty easily. Furthermore, sin(pi) = 0, so can I just ignore the pi in sin(pi + 3pi/8)?

If not, must I use some combination of the addition and half angle identities? I need to be careful if this is the case... I'd certainly be willing to give that a try on my own, but I'm wondering if I'm on the right track.

Thank you.

P.S. - This thread is fun. I plan to give back later when I'm more capable. Thanks a lot, everyone.

Just a few more comments/concerns about this statement. This is not really how Trig Functions work, is it? I guess with such little practice, it's still a little tough to wrap my head around these concepts. I mean, just because sin(pi) = 0, I can just subtract multiples of pi from inside the parenthesis, right? Coincidentally, sin(pi + pi) = sin (pi)... but OTOH, sin(pi/2) != sin(3pi/2) <=> 1 != -1. BUT, the absolute values are equivalent... and if I'm not mistaken, that should always hold? Hmm...

I'm particularly interested in some clarification on the above quoted question.

TYVM!

On phone so not quoting but

* you can't ignore the pi. Why? Sin is 2pi periodic, not pi. Look at the graph of sin(x). Can you translate by pi without changing the value? (no) Algebraically, if you apply the sum formula, you'll get a cos(pi) not just a sin(pi).

* you should be able to show your "absolute value" claim with the sum formula. Try writing it up...

Imo, most of these problems to simplify trig expressions have very little value, other than familiarizing yourself with some algebraic manipulation and identifying a sum or product.

Focusing more on the properties of the functions, their applications: for example: uses in series expansions of periodic functions, or what equations they are solutions to is more important.

Look up how doing a variable transformation, you can change any of these "prove it holds" expressions into a ratio of polynomials of one variable.

Wow have I not done trig in a while, but I think you want these:

sin(x) = sin(Pi - x)
sin(–x) = –sin(x)

Rather than learn these by memory, just look at a sine graph and think about what these identities are saying. They're just based on the symmetry of the graph, and if you see why they're true, you'll be able to produce them on the fly without needing to memorize a thing. You'll also be able to work out what sin(11Pi/8) is equivalent to (by which I mean, you'll be able to see what other angles produce the same sine, or the negative of that sine) without really doing any work, just looking at the periodic and reflection symmetries of the sine graph.

Or just proceeding algebraically using the above identities you get:

sin(11Pi/8) = sin(Pi - 11Pi/8) = sin(-3Pi/8) = -sin(3Pi/8)

Of course you then need to do some actual work to evaluate that.

help with an eigen problem pls.

I have y'' + lamda*y = 0 , y(0) = y(2pi) , y'(0) = y'(2pi)

y = c1sin(sqrt(lamda)x) + c2cos(sqrt(lamda)x)

now how do I solve for lamda ? I just can't figure out how to do it when my initial conditions are given in this wierd way. I'm used to being given y(0) = 0 etc.

You have for the first condition

y(0)=c2
y(2pi)=c2 cos(sqrt(lambda)2pi)

If these are equal then you get 1 = cos(sqrt(lambda)2pi)

Now you can ask yourself for what values of lambda does this hold?

This isn't quite right. You actually have
y(2*pi) = c1 sin(sqrt(lambda) * 2pi) + c2 cos(sqrt(lambda) * 2pi)
But yes, setting y(0) = y(2pi) gives you an equation relating c_1, c_2 and lambda. The same is true if you compute y'(x) and then set y'(0) = y'(2pi). Only certain values of lambda will allow you to find a solution to both equations, and you have to work out what they are. (Hint: combine the two equations to get a single one with just lambda and no c_1 or c_2).

I'm writing a paper where I compare unemployment rates and salary levels between Americans with high school diplomas and the average American, and I am wondering what the most appropriate way to compare the figures are.

If unemployment rate for non diploma workers is 14.1% and the national average is 7.6%, then is it appropriate to say that those without diplomas are 85% more likely than the average American to be unemployed: ((14.1-7.6)/7.6)*100=85

Similarly if those without high school diplomas make $451 and the average American makes $797, do we say that the non diploma group makes 43% money than the other group? ((451-797)/797)*100=43

Or is there a better way to discuss these types of numbers?

Thanks.

Hey good catch there. I shouldn't try to post help at 1am!

Intuitively, it makes sense to me. And when I look at a point on the unit circle, and look at the point pi distance away from it (in radians), it makes sense too.

But when I look at the sum formula, I'm not sure how to begin to prove this algebraically.

sin(a + b) = sin(a)cos(b) + cos(a)sin(b)

Hint please?

I've had some ideas, but they all seem to take me off on some tangent.

I'm glad to hear this, because this has been considerably tougher than previous coursework, and I think I'm one of the better students in the class. I Googled variable transformation with no luck.

As stated above, I'm not sure how to derive this, but it makes sense to me when looking at the unit circle. Of course being able to derive a formula greatly increases retention, and eventually I plan to be able to! But since I'm still learning and struggling with this, I'll work through some examples first. I'm going to repeat the steps you posted just to work through it on my own.

Evaluate: sin(11pi/8)

Using the identity: sin(x) = sin(pi - x)

sin(11pi/8) = sin(pi + 3pi/8) = sin(pi - (-3pi/8)) = sin(-3pi/8) = -sin(3pi/8)

Now, using half angle identity, I will first solve this problem for +sin(3pi/8), then multiply the result by -1.












Now since sin[(3pi/4)/2] is in the first quadrant, we would select the positive root, but then I'm multiplying the result by -1, since I want -sin[(3pi/4)/2]. So, the answer is:



Thanks, everyone!

a=pi

Take a look at the section "Rational Forms of Common Curves" at http://www.cs.mtu.edu/~shene/COURSES.../rational.html

-2Pc + X(1-Pc)=0

was reduced to:

Pc= X/(2+X)

So first we would go:

-2Pc + X - XPc=0

what are the rest of the steps to get "Pc = x/(2+x)?"

Well you clearly need to get one Pc term, so can you figure out a way to do that?

It is one term. In the example, Pc=Probability. Why it wasn't just P, I don't know (edit: just watched the video again, the "c" represents the column player in the matrix, that's all).

You have 2 terms above with Pc. You need to get one to solve for it.

You have one with an 'X' multiplied by it and one with a "-2" multiplied by it.

If you want to get a formula, Pc = blah blah, first you need to get all the Pcs on one side of the equal sign, alone.

that's exactly what I'm asking how to do

Now, subtract X from both sides.

That'll make the formula

-2Pc - XPc = -X

Then since these are all negative, I would multiply it all by -1, so:

2Pc + XPc = X

Then try to work it from there. Do you see the remaining steps to reduce it to: Pc= X/(2+X) ??

All the hints in the world might not help here if 28renton isn't familiar with the algebraic techniques. So I'll give a more direct hint. If you had an equation like:

ab + b = c - ac

and wanted to solve for a, you would:

* first get your terms with a to one side, and the other terms to the other side:

ab + ac = c - b

* factor the 'a' from the sum on the left side (this is the key step) :

a(b+c) = c - b

* divide on both sides by the expression in brackets on the left side:

a = (c - b) / (b + c)