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It is so consistent because mathematicians have worked hard to make sure it is. Nobody understands quantum mechanics *

Whizzies wrote:

The compound angles is something that I have to learn (should know it a lot better than I do now) too. I want to know why it is like that and not just put it in the TI-84 and be done with it.

sin(30) = 0.5 and sin(60) = 0.866, but sin(90) is not sin(30) + sin(60). Functions which have the property f(a+b) = f(a) + f(b) are called linear. If you draw the graph you get a straight line. Sine doesn't behave like that and its graph isn't straight. So how do you calculate sin(a+b) if you know sin(a) and also sin(b) ? Turns out you have to know cos(a) and cos(b) as well.

You'll find a bit about the compound angle formulas at the bottom of this page:

http://www.mathsisfun.com/algebra/trigo … ities.html

How do you prove them? I know a geometric proof that is OK for acute angles. There's a general proof that uses matrices. You have to know how to multiply matrices and what the general form of a rotation matrix is.

Whizzies wrote:

I know that people that work with land use a different kind of ... angles ? Like I believe they have circles of 420 degrees? Don't know.

I've never heard of this. Did you mean gradians? http://en.wikipedia.org/wiki/Gradian

Mathematicians also use a measure called the radian. It is more widely used once you get beyond school level math.

Whizzies wrote:

So the unit circle is a more general theorie of the sine, cosine and tangent? Would you say that the ''normal sine, cosine and tangent'' are incorporated in the unit circle?

Yes and yes.

Bob

*

Feynman wrote:

"If you think you understand quantum mechanics, you don't understand quantum mechanics."

Ok. Extra bits on diagram:

h = HD = CI. The angles marked with a dot are equal (alternate angles). Say, alpha.

So using the area I have already worked out and

where a and b are the lengths of the parallels, you can work out FG.

Bob

hi harrychess,

By using Sketchpad and measuring, I'm getting this:

AC = 10, BD = 8. AB parallel to DB.

I can move E about, and still get the same result for FG, which is interesting as the whole shape changes.

Using those right angled triangles, it's not too hard to show that the area of ABCD = 0.5 x 8 x 10.

So if I could calculate h, the height of the trapezium, then the length of FG would be easy.

But I cannot see how to get h just yet. Still thinking ....................................

Bob

oscillating: eg. 1, -1, 1, -1 ,1, -1, .............

periodic: eg. 1,2,3,5,1,2,3,5,1,2,3,5, ..........

I suppose you could argue that oscillating sequences are a special case of periodic.

Bob

Did you mean

When I first learnt about convergence, the implication was that if it didn't converge then it diverged. But, more recently, I have met a text that listed four possibilities:

converge, diverge, oscillate and periodic.

Putting a name to it, is not as important as describing what the sequence is doing.

Bob

hi headhurts,

headhurts wrote:

Ability to measure vertical angles

Do you mean with something like this:

http://www.instructables.com/id/Using-a … re-height/

Bob

hi Whizzies

Mathematics is concerned with building 'models' of the real world (that are useful). A theory starts with some definitions and axioms, and then theorems are constructed from these. A model has to be consistent (eg. In one place you might prove that pi = 3.14.... You mustn't prove in another part of the theory that pi = 49. That would be inconsistent.) After that a model can be pretty much anything. If it's useful, that's a bonus.

For years a lot of number theory to do with primes was thought to be the concern of only pure mathematicians; ie. not generally useful. But with the advent of the internet, prime number theory has become really useful as it is used for encryption of passwords.

The trig. ratios were first dreamed up as ratios in a right angled triangle (sine = opposite/hypoteneuse etc) but then somebody thought it would be useful to have a more general definition so that sine(240), for example, also has a meaning.

If you start with a pair of coordinate axes and draw on a circle, centred on the origin, you can make a new definition for sine that has something to do with the y coordinate of a point on the circle. The revised definition has to be consistent with the old, right angled triangle definition, otherwise things that have already been proved in that case (eg. sine(30) = 0.5) won't continue to be true.

There is nothing compulsory about using a unit circle, but it makes life easier as the 'hypotenuse' is now 1.

The point starts at (1,0) and rotates anticlockwise around the origin. When it has gone 30 degrees the y coordinate is 0.5. So the new definition for sine is

If you draw a right angled triangle with hypotenuse OP, then you can see that this definition gives the same values for sine as the old definition, but now we can continue the rotation to get the sine of any angle. Cosine is defined similarly; and tangent is defined as sin/cos.

As a result we can get the sine, cosine and tangent of any angle at all, including angles over 360, and negative angles.

Lots of new properties can now be proved: for example:

This is one of a set of formulas called the **compound angle formulas**. They get used a lot.

Bob

hi anikettomar98

Welcome to the forum.

According to Wikipedia:

http://en.wikipedia.org/wiki/Logarithmi … rentiation

"where we can use it" most functions

"where we cannot use it" the article doesn't really say

"why we use it" to make tricky diffs a bit easier.

eg.

But why would you do this for an example that can be done more easily. In practice, you would use it only when it makes things easier, such as this second example:

eg2.

Bob

I think it's worse than tedious. You would have to create three right angled triangles in one diagram, and chase around all the distances. Possible, but not nice.

I cannot think of an easy way to do with a diagram.

Bob

hi Gabrielle,

8.

y = (1/2)x - 3

y = (3/2)x what sign is missing here? 1

y = (1/2)x 3

(3/2)x 1 = (1/2)x 3

+1 +1

(3/2)x = (1/2)x 4

(4/2)x = -4

2x = -4

x = -2

y = (3/2)x 1

y = (3/2)-2 1

y = -3 1

y = -4

(-2, -4)

12.

4x + 3y = -15

y = x + 2

y = x + 2 (times equation by 3)

3y = 3x + 6 ........................A

4x + 3y = -15 ..........................B

7x = -21

x = -3

y = x + 2

y = -3 + 2

y = -1

(-3, -1)

This answer is correct and it looks like you have used elimination. You can make this clear by lettering the equations and then stating what you are doing to eliminate y.

eg A - B gives 3y - 4x - 3y = 3x + 6 - (-15)

-4x = 3x +21

etc

16.

y = -2x + 1

y = x wthat sign is missing here? 5

0 = 3x + 6

3x = 6

x = 2

y = x 5

y = 2 5

y = -3

(2, -3)

Im not sure what I did wrong for these.. my teacher keeps telling me this

#8 - Tell me what you did to get from the first step to the second.

(3/2)x=(1/2)x-2

(4/2)x= -2

This step is wrong. If you subtract (1/2)x from each side you should have x = -2

#12 - The work must show elimination. You have used substitution. Try again.

#16 - I don't see your elimination work. Please look at the example that I provided in my last post to help you show your work for #12 and #16

Bob

hi,

I don't think you have to make a new diagram for this. You can use the results you already have.

The question requires you to change sin(something) into cos(something). Back in post 1 you had a formula that does that:

sin (a+0.5pi) = Yr = Xp = cos (a)

But you want to change sin(x + 1/6pi).

So put x + 1/6pi = a + 0.5pi

Now re-arrange that to make 'a' = something.

Then you can put that a into the above to get sin(x + 1/6pi) = cos a and you have your answer.

Bob

Arhhh. You say that like it is a theorem I should know. I don't recall ever meeting that one, but that's not surprising as I don't bother to learn theorems. But it was easy to prove, so I'll add it to the list of theorems that I won't bother to learn.

Then it quickly follows that the shape is a rectangle. I expect you can get XY^2 using that, but I don't think I'll bother now. I'll leave it to the interested reader.

Bob

hi demha

The rules of algebra are just the rules of arithmetic. As a result if expression A = expression B, you can substitute any value for x and you will get number = same number.

So choose a value for x and evaluate the expression in the question. Then evaluate every multi choice answer with the same choice for x. Providing you don't slip up with the calculations, if one multi choice has the same result as the question, that must be the answer.

But there is a small chance that your choice for x will lead to two multi choice answers giving the same, correct result. If that happens you would have to make another choice for x to determine which answer is right.

eg.

5. x^9 + 1

A (x + 1)(x^2 - x + 1)(x^6 - x^3 + 1)

B(x^3 + 1)(x^6 + 1)

C(x + 1)(x^2 - x + 1)(x^3-1)2

D(x + 1)3(x^2 - x + 1)3

E (x^3 + 1)(x^6 - x^3 + 1)

Choose x = 1

Question = 1^9 + 1 = 2

A = 2 x 1 x 1 = 2

B = 2 x 2 = 4

C = 2 x 1 x 0 = 0

D = 2 x 3 x 1 = 6

E = 2 x 1 = 2

So we have two contenders: A and E

choose another x, say, x = 2

Question = 512 + 1 = 513

A = 3 x 3 x 57 = 513

E = 9 x 57 = 513

Oh dear. Could this mean both answers work?

It would if x^3 + 1 = (x+1)(x^2-x+1).

I'll leave it to you to show that this is indeed true.

So which answer should you pick?

You'll have to 'read the small print'. Does it say 'completely factorise' ? If so then pick A. If it doesn't, I'd still pick A as a superior factorisation. But, if this is CompuHigh, it could just be that the question compiler made a mistake.

Bob

Thanks Stefy.

Bob

It is also the case that cos(-a) = cos(a) for any angle a. Can you prove this ? So this last result follows from that.

Bob

hi Whizzies

cos (a-0.5pi)=sin(a)

For this you need a diagram like mine, but with the rotation going clockwise rather than anticlockwise.

The 'up' for angle a will become the 'across' for a - 0.5pi in the fourth quadrant.

cos (a-0.5pi) = Qx = Py = sin (a).

is correct.

what is the difference between cos ( pi - a); cos (a - pi) ?

They are the same. Again a diagram will show this. Start with an angle a. Rotate by -pi. This takes the angle from the first into the third quadrant. Now start with angle pi and rotate by -a. This also takes you to the same angle in the third quadrant.

Bob

When P is rotated pi/2, that triangle rotates by that amount. The across distance becomes an 'up' distance and the up distance become a negative across distance.

So the Xr coordinate is the same as the Yp coordinate but is now a negative amount

and

the Yr coordinate is the same as the Xp coordinate.

Remember these are just numbers, and those formulas are telling you how one set of numbers is related to another.

So Xr = -Yp and Yr = Xp.

If Yp = 0 (the point MUST be on the circle) so P is the point (1,0). When it rotates pi/2 to point R this is (0,1). So 0 = -0 and 1 = 1. No inconsistency.

Hope that helps,

Bob

Did you try rotating the point on the circle?

B

hi Whizzies

Welcome to the forum.

Take a look at this page:

http://www.mathsisfun.com/geometry/unit-circle.html

A short way down the page, you'll find an interactive circle that will help to answer all your questions.

A similar version with graphs is at

http://www.mathsisfun.com/algebra/trig- … ircle.html

Bob

hi jjoy

Welcome to the forum.

The algebra is messy but it should work. Sorry, I cannot find a simpler method.

Two tangents to a circle from a point will be equal.

Then the small (removed) cone will be in the same proportion as the large cone. Hence:

I don't have a simple solution to these, but someone else on the forum will probably put the equations through a computer for us and come up with values for x and y.

Then it should be easy (?) to compute the volume.

Bob

Stefy wrote:

Well, the opposite sides are equal and parallel...

From my diagram and analysis YY' is parallel to XX' (as both are perpendicular to BD) and X'Y = XY' (by reflection). I don't yet have evidence that X'Y is parallel to XY' nor that X'X = YY'.

But I have simplified my calculations:

Let BCA = a and XCY = b

By simple trig cos(a)=12/13 and sin(a) = 5/13

Therefore, cos(b) = cos(a+30) = cos(a)cos(30) - sin(a)sin(30)= (12root3 -5)/26

Then XY^2 = 144x3 +169/4 -6root3(12root3-5) = 310.2115

This is confirmed by my Sketchpad drawing.

Bob

It is my hope to prove it is a rectangle. So far I have only proved the isosceles trapezium. I am about to tidy up and finish off my method. It will take a little while.

What this space.

Bob

A better version.

Because YD = 12 and CD = 24, that makes CYD a 30-60-90 triangle.

Angle ACB can be found thus:

cosine ACB = 12/13 and thus sine ACB = 5/13.

cosine BCD = cosine(BCA + 60) = cosBCA.cos60 - sinBCAsin60 = (12/13 - 5root3/13)/2 = (12 - 5root3)/26

***find sine BCD. Running out of time now, I'll fill this in later.

cosine XCY = cosine(BCD - 30) = cos BCD.cos30 + sinBCD.sin30

Then use cosine rule on XCY.

Note: I've used compound angle formulas to avoid the 'inaccuracy' of using a calculator for these angles.

Bob

OK. Like this:

AC = 24. Use cosine rule to calculate ABC and hence BCD. Subtract 30 and you have YCB. Then use cosine rule on XYC.

Checking ..........................

Bob

hi Stefy,

Yes, you're right. The above is flawed. I'm working on a corrected version. At the moment X'YY'X is an isosceles trapezium. I'm sure I had it sorted a while back but I cannot re-trace my steps. I had a way of showing XZ = ZY and that forces the trapezium to become a rectangle. Then you probably have to use cosine rule.

Also DYC = 90. This should help but I need angle YCX.

Thinking still ........................................

Bob