2

I am reading "What Is Mathematics? An Elementary Approach to Ideas and Methods" And I am stuck here, I don't get it. I have posted a screen shot underlining what my doubt is..

I dont get it when the author says while the pythagoras theorem is : $a^2 + b^2 = c^2$ and then he says $x=a/c$ and $y=b/c$ and then the equation should be according to me , $ax+by=c$.. right?? but the author writes y^2 = (1-x)(1+x)

Which I think may have came from something like $x^2 + y^2 = 1$ $y^2= 1^2 + x^2$ $y^2= (1+x)(1-x)$ (since $a^2 + b^2 = (a+b)(a-b)$ )

but I dont get it where did that x^2 + y^2 = 1 came from ?? Is it that author assumed that x=a/c and y=b/c and then stoped talking about pythagoras theorem and started talking on x^2 + y^2 = 1???

and then further he introduces a number t.. i don't get it how it got converted into y=t(1+x) and (1-x)=ty ???

Can please someone help?

also , while i was writing the question it clicked me that if: x=a/c (i.e. opposite upon hypotenuse means x is sin ) y=b/c (i.e. adjacent upon hypotenuse means y is cos) therefore x^2 + y^2 = 1 (sin^2 + cos^2 = 1)

but then if i assume i am right about the sin cos thing then how come the last step is derived ?? (the one in blue color underline and box) also that as per me x is sin , but the formula in that book is of cos2x? when t=tanx??

so m lil confused .. if you want any more clean way of me asking my doubt then do tell me i will rephrase the entire question .. :)

Image of my question asked above,, screenshot of the book

Davide Giraudo
  • 172,925
Mihir Solanki
  • 397

2 Answers2

2

For the first part of your question: the author divides both sides of the Pythagorean equation by $c^2$. This yields: $\frac{a^2}{c^2} + \frac{b^2}{c^2} = 1$. Now, he defines $x = a/c$ and $y = b/c$. The equation can than be rewritten to: $x^2 + y^2 = 1$. Substracting $x^2$ from both sides yields $y^2 = 1 - x^2$, or equivalent: $y^2 = (1-x)(1+x)$.

He then devides both sides by $y$ and $(1+x)$. This gives $y/(1+x) = (1-x)/y$. His next step is to say that the expression is actually $t = t$, where $t$ is a rational. So, $ y/(1+x) = t \rightarrow y =t(1+x)$ and, in the same way, $(1-x)/y = t \rightarrow (1-x) = ty$.

Working this out gives the two equations $tx - y = -t$ and $x + ty = 1$. Now you have two unknowns in two equations.

This can be easily solved:

$tx - y = -t \rightarrow x = \frac{y-t}{t}$. Substitute this into the second equation. That gives us$\frac{y-t}{t} + ty = 1 \rightarrow (y-t) + t^2y = t$. Now, solve for y: $y + t^2y = 2t \rightarrow y(1+t^2) = 2t \rightarrow y = \frac{2t}{1 + t^2}$.

Substitute this value for y back into the first to get the expression for $x$.

Olivier
  • 171
  • 8
  • well i reached till y=2t/(1+t^2) ... But when i try to substitute it in tx-y=-t ... it forms , tx-2x/(1+t^2)=-t .. how should i simplify it >?? Cuz in my acedemics we were never taught to take LCM of a number like 1+t.. but always for single variable or single constant.. – Mihir Solanki Nov 04 '13 at 17:39
  • also how are you writing LaTeX ?? do you use some program or u just know the code for writing it? – Mihir Solanki Nov 04 '13 at 18:22
  • The substition is fairly easy:

    $x = \frac{y-t}{t} = \frac{\frac{2t}{1+t^2} -t}{t} = \frac{2t -t -t^3}{t(1+t^2)} = \frac{1 - t^2}{1 + t^2}$. Also, you can type most of the usual Latex code if you type it inbetween dollar signs.

     – Olivier Nov 04 '13 at 20:17
  • In that x=[2t/(1-t^2) -t]/t when taking lcm .. u multipled the t by (1+t) ??? – Mihir Solanki Nov 05 '13 at 02:59
  • Got it .. I Actually tried multiplying the whole term 1+t^2 to t .. but then I obtainted a t^3 on simplyfying halfway .. so I thought I was wrong but forgot about the t in the denominator .. :p – Mihir Solanki Nov 05 '13 at 03:13
1

You replaced one of the two $a$s making up $a^2$ with $cx$. Try replacing both (and similarly for the two $b$'s in $b^2$ with $cy$)

draks ...
  • 18,449
Hagen von Eitzen
  • 374,180