Trigonometry proof of formula

Whizzies · 2014-07-26 23:33:25

Heyy, I am unable to upload a photo so I cannot refer to one.

But I am working on understanding Trig formulas like: sin(-A) = -sin(A).

the book tries to explain it by portraying two unity circles (ill descibe them both).

First figure is a unit circle with a point P in the first quadrant and a point Q in the fourth (those two dots are mirrorring). The angle is called a)

Point Q is the mirror off P in the x-axis so the angle of rotation of Q is -a and Xq = Xp and Yq = -Yp.
So
sin (-a) = Yq = -Yp = -sin(a) and
cos (-a) = Xq = Xp = cos(a)

Yq and Xq the q is meant a s subscript and the Y and X are just normal y and x (sorry I dont know how to subscript).

My question is why is it!! sin (-a) = Yq = -Yp = -sin(a) it should be positive and Yq should be negative right? Why are they turned? I know you could manipulate the equation Yq = -Yp, but that would just make it more complicated
figured it out! After staring at it for half an hour!

Second question is. a unit circle with Point P (quadrant I, and same spot as the first question) and point R (quadrant 2 and is (a+0.5pi) from point P).

Point R is the image of point B with the roation of 0.5pi rad. Xr = -Yp and Yr = Xp.
So
sin (a+0.5pi) = Yr = Xp = cos (a)
cos (a + 0.5pi) = Xr = -Yp = -sin(a)

I don't uderstand the expression that I made BOLD. Why is Xr = -Yr? Sorry that I am unable to opload the picture so that it would have been easier to explain?

Last edited by Whizzies (2014-07-27 00:04:37)

Bob · 2014-07-26 23:54:10

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

Whizzies · 2014-07-27 00:00:11

bob bundy wrote:

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.
Bob

Hé (: I have been there before. I understand the ''formulas'' in the normal x,y-functions and how to make the cosine into sine and vica versa, but with the unit circle it is a different story (:

Bob · 2014-07-27 00:03:40

Did you try rotating the point on the circle?

B

Whizzies · 2014-07-27 00:06:19

bob bundy wrote:

Did you try rotating the point on the circle?
B

Yep, but I don't understand this:

Xr = -Yp and Yr = Xp

I also changed my first post! (:

I don't see how Xr = -Yp cause how can X be Y (: And if Y equals zero then X is also zero that would mean it is the origin.

Bob · 2014-07-27 00:54:31

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

Whizzies · 2014-07-27 03:47:11

bob bundy wrote:

http://i.imgur.com/udhAc20.gif
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

WOW! I just... I did not see it that way! it helped a lot! I just... well it is correct; math stills amazes me sometimes how it just 'fits together'! I thank you very much!

btw, I was looking at the coördinates and not the triangle it self (: But when I look at the triangle I understand the .. unity!

Last edited by Whizzies (2014-07-27 04:12:47)

Whizzies · 2014-07-27 04:42:26

hé I had to ''proof'' cos (a-0.5pi)=sin(a) (sorry no picture)

Does this 'fit' I had to figure this one out myself because the anwer book has not a explained answer just how it ''is''.

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

Point P is in the first quadrant and point Q is in the fourth. If this is correct, I think I get the point! (:

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

Last edited by Whizzies (2014-07-27 04:52:45)

Bob · 2014-07-27 05:31:38

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

Bob · 2014-07-27 05:33:55

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

Whizzies · 2014-07-27 06:27:41

bob bundy wrote:

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

YES! I am very very happy that I have it right (: Means that I am on the right way!

bob bundy wrote:

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

cos (-a) = Xq = Xp = cos (a)

I now have to do sin(x + (1/6)pi) = cos (ax + b)

I have to learn these formulas -sin(a) = sin (a +pi) and sin^2(a) + cos^2 = 1 I have 8 bases. Just two are derived and proven, but I really want to understand what I am doing so I have to figure it out. And thanks to this forum and you! I am succeeding (:(:(: I am happy that I found this forum!

First you have to take sin (a+0.5pi) = Yr = Xp = cos (a) and then .. yep still needs some thinking!

PS I am drawing circles and lines to figure it out (:

Bob · 2014-07-27 07:04:47

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

Whizzies · 2014-07-27 22:35:28

Thx (:

Would it also be possible to make a diagram here, or is that tedious work?

Bob · 2014-07-27 23:50:44

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

Whizzies · 2014-07-28 22:34:35

bob bundy wrote:

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

Yes, I tried to do it but I got confused so I gave up and asked you if it were possible and your answer explains my confusion (:

Could you also explain maybe what the unit circle is exactly? I know the ''drill'' but what is it? If I were to have a right triangle (the special ones with the nice rations) then I would understand how handy it is.

I once saw on youtube how triangles are so nice! They support bridged and make you able to measure up distances. I believe even on schips (in the old days) they had a telescope that measures things with triangles, but what is the use of the unit circle?

Bob · 2014-07-29 00:24:50

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

Whizzies · 2014-07-30 08:38:03

I have read it (: I will respond tomorrow (: (:

Last edited by Whizzies (2014-07-30 08:38:12)

Whizzies · 2014-07-31 01:47:58

Sometimes I am awestruck by mathematics, I just can't understand how it can be so consistent! Sometimes (even with quantum mechanics (what I definetely don't understand) I sometimes think that we create the reality, because math is build of axioms and definitions. I don't know... for now I will stay awestruck and hope I will be as good as the people on this forum (:

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. I understand the unit circle a little bit better now, but still triangles and same points even though it looks different, still is a little bit...

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.

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?

Last edited by Whizzies (2014-07-31 01:48:59)

Bob · 2014-07-31 03:20:32

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."

Whizzies · 2014-08-01 01:56:36

I would recognize a matrice, but I wouldn't know what it is. I will be learning about the compound angles very soon also the formules from Mollweide will I be studying, but I am a bit in time distress!

Radian is not radians? I did mean Gradients, yes! Haha I thought it was 420! But I never used it or have read properly about it. I know that I ask sometimes question that are already explained on the website, but sometimes I am too uncertain if it really is like that (:

Before I started to delve a little bit deeper in the unit circle; I always thought that it were to seperate things, but evidently they are connected with eachother. I still don't understand properly what the function precisely is. I just have to go a little bit deeper and try to make connections and such (:

Feynman did say that, did he not! But I guess one day there will be someone who understand it, and knows it very well! I have a small cursus of the special relativity of Einstein, that I want to absorb, but my teacher said that it would be wiser if I first strenghten my mathematic abilities (:

Math Is Fun Forum

#1 2014-07-26 23:33:25

Trigonometry proof of formula

#2 2014-07-26 23:54:10

Re: Trigonometry proof of formula

#3 2014-07-27 00:00:11

Re: Trigonometry proof of formula

#4 2014-07-27 00:03:40

Re: Trigonometry proof of formula

#5 2014-07-27 00:06:19

Re: Trigonometry proof of formula

#6 2014-07-27 00:54:31

Re: Trigonometry proof of formula

#7 2014-07-27 03:47:11

Re: Trigonometry proof of formula

#8 2014-07-27 04:42:26

Re: Trigonometry proof of formula

#9 2014-07-27 05:31:38

Re: Trigonometry proof of formula

#10 2014-07-27 05:33:55

Re: Trigonometry proof of formula

#11 2014-07-27 06:27:41

Re: Trigonometry proof of formula

#12 2014-07-27 07:04:47

Re: Trigonometry proof of formula

#13 2014-07-27 22:35:28

Re: Trigonometry proof of formula

#14 2014-07-27 23:50:44

Re: Trigonometry proof of formula

#15 2014-07-28 22:34:35

Re: Trigonometry proof of formula

#16 2014-07-29 00:24:50

Re: Trigonometry proof of formula

#17 2014-07-30 08:38:03

Re: Trigonometry proof of formula

#18 2014-07-31 01:47:58

Re: Trigonometry proof of formula

#19 2014-07-31 03:20:32

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#20 2014-08-01 01:56:36

Re: Trigonometry proof of formula

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