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Sine ecuations

  1. Jan 3, 2008 #1
    Say we have the ecuations below a,b,c positive numbers
    sin(A)=a/sqrt(a^2+b^2) 1
    sin(B)=a/sqrt(a^2+c^2) 2
    sin(A)cos(A)=sin(B)cos(B) 3

    I realised that calculating cos(A) and cos(B) and then solving the third ecuations i get the solutions b=c and a^2=bc
    But solving only sin(A)=sin(B) ( we get form 3 that A=B ) I get only b=c . I'd like to know where did the second solution go .... ??
     
  2. jcsd
  3. Jan 3, 2008 #2
    did anyone suggested you of latex format
     
  4. Jan 3, 2008 #3
    Nobody ( just) did
     
  5. Jan 3, 2008 #4
    [tex]sin(A)\ =\ \frac{a}{\left(a^{2}+b^{2}\right)^{\frac{1}{2}}}[/tex]

    [tex]sin(B)\ =\ \frac{a}{\left(a^{2}+c^{2}\right)^{\frac{1}{2}}}[/tex]

    [tex]sin(A) \times cos(A)\ =\ sin(B) \times cos(B)[/tex]

    Did it for ya :P
     
  6. Jan 3, 2008 #5
    Tks. I shall learn this latex .
     
  7. Jan 3, 2008 #6
    It is not really true that b=c, or that A=B. They could be equal, but there are many solutions where they are different, for example a=sqrt(3), b=1, c=3, A=60 deg, B=30 deg.

    If you draw a plot of the function f(x) = sin(x)cos(x), it could help you to visualize the relation between A and B (and thus between b and c).

    Also, since a,b,c > 0, you can draw them as line segments, so it might help to draw a picture with triangles that represent eq. 1 and 2 (and where b is not necessarily equal to c).
     
    Last edited: Jan 3, 2008
  8. Jan 3, 2008 #7
    one thing is clear seeing the eq. they certainly represent the right triangle with a perpendicular dropped on hypotensue

    a=length of perpendicular
    b,c sides

    i hope their is no specific solution for the angles

    they are just complementary

    don,t go by it it is just a guess , no calculations
     
    Last edited: Jan 3, 2008
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