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Derive the Pythagoream Identity

  1. Aug 6, 2009 #1
    1. The problem statement, all variables and given/known data

    Derive sin^2 + cos^2 = 1


    2. Relevant equations
    Use cos 0 =1, cos (x+y) = cos x cos y - sin x sin y

    Earlier someone posted this same question, but I still dont understand it so please help
     
  2. jcsd
  3. Aug 6, 2009 #2
    [tex]\sin A = \frac {\textrm{opposite}} {\textrm{hypotenuse}} = \frac {a} {h}\,.[/tex]
    [tex]\cos A = \frac {\textrm{adjacent}} {\textrm{hypotenuse}} = \frac {b} {h}\,.[/tex]

    [tex]sin^2A=\frac{a^2}{h^2}[/tex]

    [tex]cos^2A=\frac{b^2}{h^2}[/tex]

    [tex]sin^2A+cos^2A=\frac{a^2+b^2}{h^2}=[/tex]

    From Pythagorean theorem [itex]h^2=a^2+b^2[/itex] then

    [tex]=\frac{h^2}{h^2}=1[/tex]

    Regards.
     
  4. Aug 6, 2009 #3
    This is essentially one of the arguments outlined in the previous thread. For any real number x, we can apply the identity cos(a - b) = cos(a)cos(b) + sin(a)sin(b) for real a,b.

    1 = cos(0) = cos(x - x) = cos(x)cos(x) + sin(x)sin(x).
     
  5. Aug 6, 2009 #4

    HallsofIvy

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    Just out of curiosity, how would you prove cos(a- b)= cos(a)cos(b)+ sin(a)sin(b)? All proofs I have seen rely on the Pythagorean theorem in some form.

    And, mathismyworld, please don't post the same twice, especially under different names!
     
  6. Aug 6, 2009 #5
    Maybe by playing around with the Maclaurin polynomial, you could "define" cosine that way for cos(a-b) and avoid having to rely on the Pythagorean theorem?
     
  7. Aug 6, 2009 #6

    zcd

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    This is for the limited case of [tex]0<\theta<\frac{\pi}{2}[/tex], but draw out a right triangle with angle [tex]\theta[/tex] and hypotenuse of length L. The side opposite to [tex]\theta=L\sin(\theta)[/tex] and the side adjacent to [tex]\theta=L\cos(\theta)[/tex]. Applying the Pythagorean theorem will get you [tex]L^{2}=L^{2}\sin^{2}({\theta})+L^{2}\cos^{2}({\theta})[/tex]. This simplifies to [tex]1=\sin^{2}({\theta})+\cos^{2}({\theta})[/tex], the Pythagorean identity.
     
  8. Aug 6, 2009 #7
    I was curious about this too and did a google search. The http://www.trans4mind.com/personal_...metry/basicFormulae.htm#Compound_Angle_Proof" I found proved the compound sine formula first using a triangle with an angle of [tex]\alpha + \beta[/tex] and the formula for the area of a triangle [tex]A = \frac{1}{2} ab \sin(C)[/tex]. Then to get cosine, you'd use the identity that [tex]\cos(x) = \sin(\frac{\pi}{2} - x)[/tex]. I don't believe either of those rely on the Pythagorean Identity, but at this hour of the night, I'm not sure whether this would be enough to conclude that this would work for any angle.
     
    Last edited by a moderator: Apr 24, 2017
  9. Aug 6, 2009 #8

    rock.freak667

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    http://img30.imageshack.us/img30/6302/pictq.jpg [Broken]


    Crude image but here goes (according to my text book):

    THe right angled triangles OPQ and OQR contain angles A and B as shown. The dotted lines are construction lines and the angle URQ is equal to A.

    [tex]sin(A+B) \equiv \frac{TR}{OR} \equiv \frac{TS+SR}{OR} \equiv \frac{PQ+SR}{OR}[/tex]


    [tex]sin(A+B) \equiv \frac{PQ}{OQ} \times \frac{OQ}{OR} + \frac{SR}{QR} \times \frac{QR}{OR}[/tex]

    [tex]sin(A+B) \equiv sinAcosB +cosAsinB[/tex]

    replace A by [itex] \frac{\pi}{2}-A[/itex] tp get cos(A+B)
     
    Last edited by a moderator: May 4, 2017
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