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Basic Trigonometry

  1. Sep 1, 2008 #1
    1. The problem statement, all variables and given/known data
    If [tex]m^{2}+m' ^{2}+2mm' cos \theta=1[/tex]

    [tex]n^{2}+n' ^{2}+2nn' cos \theta=1[/tex]

    [tex]mn+m'n'+(mn'+m'n)cos \theta =0[/tex],
    then prove that [tex]m^{2}+n^{2}=cosec^{2} \theta[/tex]




    3. The attempt at a solution
    I have tri a lot but its difficult to eliminate m' or n'

    If I add the first to equations and separate m^2+n^2 to left ide, again its difficult to prov the RHS to [tex]cosec ^{2} \theta[/tex]
     
  2. jcsd
  3. Sep 2, 2008 #2

    tiny-tim

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    draw the triangle!

    Hi ritwik06! :smile:

    Well, looking at the the first two equations, you're obviously supposed to draw a triangle with sides m and m', and an angle of π - θ between them. :wink:

    Give the name φ to the angle opposite m (so the angle opposite m' is … ?).

    Then do the same with n n ' and ψ.

    And then use the sine rule … :smile:
     
    Last edited: Sep 2, 2008
  4. Sep 2, 2008 #3
    Re: draw the triangle!

    Hi tim :smile:

    Well I am not familiar with this. though I have used sine rule in physics. Please tell me why am I expected to make a triangle ?? and what shall I do with this thing? Simply applying:

    sin a /a=sin b/ b =sin c /c wont solve my purpose, i have tried it. :wink:
    regards
     
  5. Sep 2, 2008 #4

    tiny-tim

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    cosine rule for triangles …

    ah … perhaps you haven't done the cosine rule for triangles …

    a2 = b2 + c2 - 2bc cosA​

    so the reason to make this particular triangle is because the first two equations given are the cosine rule for two triangles, both with one side of length a = 1, and both with opposite angle π - θ, one with the other sides of length m and m', and the other n and n'.

    (check for yourself that the cosine rule really does give those two equations for those triangles)

    Draw those triangles, and apply the sine rule to get m and m' in terms of θ and φ, and n and n' in terms of θ and ψ.

    Then plug those values into the third given equation, which should give you a nice relationship between θ φ and ψ. :smile:
     
  6. Sep 2, 2008 #5

    HallsofIvy

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    Re: draw the triangle!

    Because the cosine law says that if a triangle has sides of length a, b, c and A is the angle opposite side A, then c2= a2+ b2- 2ab cos(A).

    The equation [itex]m^2+ m'^2+ 2mm'cos(\theta)= 1[/quote] differs from that only only in having "+" instead of "-". Since [itex]cos(\pi- \theta)= -cos(\theta[/itex]), that should make you think of a triangle having sides of length m, m', and 1 and having angle [itex]\pi- \theta[/itex] opposite the side of length 1.
     
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