x/y = infinity ? Plz.. help

[I am a brawl guy]

Ok!
So today I spend my whole day thinking about it:

We all know
Xcos(alpha)+ysin(alpha)=p ----equation (I)
right?
Then first look this:
and say
Can the :
Xcos(alpha)+ysin(alpha)=p
Be said as
X [OM]/[OA]+y [OM]/[OB]=P ?
IF YES GO TO NEXT STEP
IF NO REPly THE REASON

and in triangle OMA
sin(beta) = OM/OA WHICH IS ALSO EQUAL TO COS (ALPHA)
and in triangle OMB
Cos(beta) = OM/OB WHICH IS EQUAL TO SIN(ALPHA)

NOW SUBSTITUTE THE VALUES OF COS AND SIN ALPHAS TO SIN AND COS BETAS RESPECTIVELY

Then we get this:
Xsin(beta)+ycos(beta)=p ----equation(ii)

Now subtracting equation I and ii we will get this :


Now we know
As I have already said
Up

Sin(alpha) = cos(beta
And
Sin (beta) = cos (alpha)

So elemating or further calculating we get this:


WHICH IS MATHEMATICALLY IMPOSSIBLE
I can't figure out of I am wrong or the maths is wrong like 1=2

 

[topsquark]

Given the equation: [math]x ~ cos( \alpha ) + y ~ sin( \alpha ) = p[/math]. (This describes an equation for y in terms of x. It is not an identity.)

(1) [math]x ( cos( \alpha ) - sin( \beta ) ) = y ( cos( \beta ) - sin( \alpha ) )[/math]

But [math]cos( \alpha ) - sin( \beta ) = 0[/math] and [math]cos( \beta ) - sin( \alpha ) = 0[/math].

So we have
[math]x \cdot 0 = y \cdot 0[/math]
You can't do [math]\dfrac{x}{y} = \dfrac{0}{0}[/math] because you would have to divide both sides by 0, which is illegal.

-Dan

Addendum: By the way, we already have the zero problem in (1).
[math]x ~ cos( \alpha ) + y ~ sin( \alpha ) = p[/math][math]x ~ sin( \beta ) + y ~ sin( \beta ) = p[/math]
When you subtract these two you already know that you get 0 + 0 = 0.

 

[I am a brawl guy]

I'll

Given the equation: [math]x ~ cos( \alpha ) + y ~ sin( \alpha ) = p[/math]. (This describes an equation for y in terms of x. It is not an identity.)

(1) [math]x ( cos( \alpha ) - sin( \beta ) ) = y ( cos( \beta ) - sin( \alpha ) )[/math]

But [math]cos( \alpha ) - sin( \beta ) = 0[/math] and [math]cos( \beta ) - sin( \alpha ) = 0[/math].

So we have
[math]x \cdot 0 = y \cdot 0[/math]
You can't do [math]\dfrac{x}{y} = \dfrac{0}{0}[/math] because you would have to divide both sides by 0, which is illegal.

-Dan

Addendum: By the way, we already have the zero problem in (1).
[math]x ~ cos( \alpha ) + y ~ sin( \alpha ) = p[/math][math]x ~ sin( \beta ) + y ~ sin( \beta ) = p[/math]
When you subtract these two you already know that you get 0 + 0 = 0.

You didn't get it
You are saying that you should multiplay by 0 at first

But I asked if you take the trigos only and arrange the x and y and the trigos you get 0/0 which is inf

Which is not the reality and
you said what I did was illegal ???
Then say there are uncountable algebra which cannot be solved without breaking the BODMAS rule

 

[topsquark]

What? All I did was apply [math]sin( \beta ) = cos( \alpha )[/math] and [math]sin( \alpha ) = cos( \beta )[/math] once it was pointed out instead of using the expression and winding up dividing by zero.

What I am saying is that if we have the equation [math]x \cdot p = y \cdot p[/math] that [math]\dfrac{x}{y} = \dfrac{p}{p} =1 [/math] unless p = 0, then there is no solution.

How is that not "getting it?"

-Dan

 

[I am a brawl guy]

What? All I did was apply [math]sin( \beta ) = cos( \alpha )[/math] and [math]sin( \alpha ) = cos( \beta )[/math] once it was pointed out instead of using the expression and winding up dividing by zero.

What I am saying is that if we have the equation [math]x \cdot p = y \cdot p[/math] that [math]\dfrac{x}{y} = \dfrac{p}{p} =1 [/math] unless p = 0, then there is no solution.

How is that not "getting it?"

-Dan

But listen as u have said [math]\dfrac{x}{y} = \dfrac{p}{p} =1 [/math]
Where both p are equal but in my equation both numerator and denominator (trigonometrical value) are not equal then

And when I asked my maths teacher he said my equation wasn't wrong but the place I have reached is wrong which is kinda a riddle I guess?

 

[JeffM]

You say

[MATH]xcos(\alpha) + ysin(\beta) = p.[/MATH]
OK. I buy it: the sum of products of real numbers is a real number.

Then you say, using measurement in degrees,,

[MATH]\beta = 90 - \alpha \implies sin(\alpha) = cos(\beta) \text { and } cos (\alpha) = sin(\beta).[/MATH]
Again OK.

Then you say

[MATH]cos(\alpha) + sin(\alpha) = p.[/MATH]
Where did that come from? It is not generally true.

 

[I am a brawl guy]

You say

[MATH]xcos(\alpha) + ysin(\beta) = p.[/MATH]
OK. I buy it: the sum of products of real numbers is a real number.

Then you say, using measurement in degrees,,

[MATH]\beta = 90 - \alpha \implies sin(\alpha) = cos(\beta) \text { and } cos (\alpha) = sin(\beta).[/MATH]
Again OK.

Then you say

[MATH]cos(\alpha) + sin(\alpha) = p.[/MATH]
Where did that come from? It is not generally true.

[MATH]cos(\alpha) + sin(\alpha) = p.[/MATH] didn't say that
I said [MATH] x sin(\beta) + ycos(\beta) = p.[/MATH]

 

[JeffM]

You do say that equation I is

[MATH]x\ cos(\alpha) + y\ sin(\alpha) = p.[/MATH]
You say that "we all know that." I have no idea what is encompassed by "that." It defines p as a function of three independent variables, but has no further meaning. Moreover, p disappears from the calculation when you subtract the yet to be defined equation II from equation I.

Then you define, using degree measurement for angles,

[MATH]\beta = 90 - \alpha.[/MATH]
From this it follows that

[MATH]sin(\beta) = cos(\alpha) \text { and } cos(\beta) = sin(\alpha).[/MATH]
[MATH]\therefore x\ sin(\beta) + y\ cos(\beta) = p \ \because \ x\ cos(\alpha) + y\ sin(\alpha) = p.[/MATH]
You call that equation II.

Now you subtract equation II from equation I and correctly reason

[MATH]x\{ cos(\alpha) - sin(\beta)\} + y\{ sin(\alpha) - cos(\beta)\} = 0 \implies[/MATH]
[MATH]x \{ cos (\alpha) - sin(\beta)\} = y\{ cos(\beta) - sin(\alpha)\}.[/MATH]
We shall call that equation III

You then make an error. You divide equation III by y, which of course you may not do unless y does not equal 0. You never stipulated that, and thus your reasoning is fallacious. But let's fix that error.

[MATH]\text {It is given that } y \ne 0.[/MATH]
[MATH]\therefore \dfrac{x \{cos(\alpha) - sin(\beta)\}}{y} = \dfrac{ \cancel y \{cos(\beta) - sin(\alpha)\}}{\cancel y} = cos(\beta) - sin(\alpha).[/MATH]
Let's call that equation IV.

You then make an another error, irredeemable this time. You divide equation IV by

[MATH]cos(\alpha) - sin(\beta) = cos(\alpha) - cos(\alpha) = 0.[/MATH]
But division by zero is not allowed in normal arithmetic.

So the whole thing is nonsense.

Now you can develop a logically consistent arithmetic where division by zero is usually allowed.

[MATH]\dfrac{a}{0} = \infty \text { if and only if } a \ne 0.[/MATH]
But that does not save you because

[MATH]x \{cos(\alpha) - sin(\beta) \} = 0[/MATH]
so your numerator is zero, which is a case where division by zero is still not allowed.

You have been told not to divide by zero. The reason is that you will get results that are nonsense.



[MATH]x{\[/MATH]

 

[topsquark]

But listen as u have said [math]\dfrac{x}{y} = \dfrac{p}{p} =1 [/math]
Where both p are equal but in my equation both numerator and denominator (trigonometrical value) are not equal then

And when I asked my maths teacher he said my equation wasn't wrong but the place I have reached is wrong which is kinda a riddle I guess?

How is that a riddle? I disagree with your teacher. If p = 0 then you can't divide by it and get a reasonable answer. It doesn't matter when you realize that p = 0. When you find out that it is you need to go back through your work and make sure you didn't do anything illegal. By even stating that the equation [math]\dfrac{x}{y} = 1[/math] without specifing that [math]p \neq 0[/math] you have made your mistake.

-Dan

Addendum: This is actually a common problem with pre-Calc and Calc I students. When they get to limits they have to start doing polynomial division. For example: [math]\dfrac{x^2 - 3x + 2}{x - 1} = x - 2[/math]. At this point you need to specify that [math]x \neq 1[/math] because it's not in the domain of the original expression. Many students don't remember to write it as [math]\dfrac{x^2 - 3x + 2}{x - 1} = x - 2 \text{, such that }x \neq 1[/math] as the full answer.